Answer (a): v₀ = mv/(M + m) (b) Rotational motion. ---------------------------- Total angular momentum before collision: Lᵢ = mvL/4 Moment of inertia of the rod before collision: Iᵣᵢ = (1/12)ML² Center of mass after collision: Let y = 0 at the bottom of the rod. Then y_cm = L/2 at the original center of mass, and y_b = 3L/4 where the ball sticks to the rod. The new center of mass is then located at: y_cm’ = (M·L/2 + m·3L/4)/(M+m) y_cm’ = (2M + 3m)/(M + m)·L/4 Moment of inertia after collision: Let d = distance between old and new center of mass. d = y_cm’ - y_cm d = (2M + 3m)/(M + m)·L/4 - L/2 d = m/(M+m)·L/4 Let r = distance between new center of mass and the point where the ball hits the rod. r = y_b - y_cm’ d + r = L/4 => r = L/4 - d r = M/(M+m)·L/4 Total moment of inertia after collision: I₀ = Iᵣᵢ + Md² + mr² where Iᵣᵢ + Md² is the moment of inertia of the rod after collision due to the parallell axis theorem.
mr² is moment of inertia added by the ball. I₀ = (1/12)ML² + M[m/(M+m)]²·L²/16 + + m[M/(M+m)]²·L²/16 … (1) Conservation of angular momentum: L₀ = I₀·ω₀ = Lᵢ ω₀ = Lᵢ/I₀ Answer (b): ω₀ = (mvL/4)/I₀ where I₀ is defined by Eqn. (1)
The CM of the ball-on-rod system is (m / M+m) * L/4 from the middle of the rod. The motion of the system is the superposition of CM translating with speed (m / M+m) * v to the right and the system rotating clockwise about the CM with angular speed (12m / 4M+7m) * v/L.
The translational motion of the pivot point after collision is to the right at speed V= v*(m/(m+M). The pivot point after collision is NOT the center of the rod. Here is my plan of attack for the rotation (which I have not completed at this time): The centrifugal force of the rod portion above the pivot plus the clay mass must equal the centrifugal force of the portion of the rod below the pivot in order for there to be no vertical translation. Use a coordinate system based at the center of the rod, and call the pivot position rp. Then calculate using moments of inertia about rp: 1) Upper portion of rod above rp: integral[(wM/L)r^2]dr, over limits rp to L/2; (check - this gives the correct value for tje upper half of a rotating rod without the clay mass) 2) Clay ball: mw((L/2)-rp); 3) Lower portion of rod below rp: same form as the part above rp, except limits are -L/2 to rp. Then use (1) + (2) - (3) = 0 to calculate rp. Fianlly, plug rp into (2) to find w. I hope this works!
To solve this problem, we need to consider both the conservation of linear momentum and the conservation of angular momentum, as well as the conditions of a perfectly inelastic collision (the ball sticks to the rod). ### 1. Linear Momentum Conservation Before the collision, the linear momentum of the system is due to the clay ball: \[ p_{\text{initial}} = m \cdot v \] After the collision, the clay ball and the rod stick together and move with a common velocity \( v_f \): \[ (M + m) \cdot v_f = m \cdot v \] Solving for \( v_f \): \[ v_f = \frac{m \cdot v}{M + m} \] ### 2. Angular Momentum Conservation We need to calculate the angular momentum about the center of mass (CM) of the rod before and after the collision. Before the collision, the clay ball has angular momentum about the CM of the rod: \[ L_{\text{initial}} = m \cdot v \cdot \left( \frac{L}{4} ight) \] Since the rod is initially at rest, it has no initial angular momentum. After the collision, the combined system (rod + ball) will rotate about the CM. The moment of inertia \( I \) of the rod about its center is: \[ I_{\text{rod}} = \frac{1}{12} M L^2 \] The moment of inertia of the clay ball about the CM of the rod (shifted by \( L/4 \)) is: \[ I_{\text{ball}} = m \left( \frac{L}{4} ight)^2 = m \frac{L^2}{16} \] The total moment of inertia \( I_{\text{total}} \) after the collision is: \[ I_{\text{total}} = \frac{1}{12} M L^2 + m \frac{L^2}{16} \] The final angular momentum is: \[ L_{\text{final}} = I_{\text{total}} \cdot \omega_f \] where \( \omega_f \) is the final angular velocity. From the conservation of angular momentum: \[ m \cdot v \cdot \frac{L}{4} = \left( \frac{1}{12} M L^2 + m \frac{L^2}{16} ight) \cdot \omega_f \] Solving for \( \omega_f \): \[ \omega_f = \frac{m \cdot v \cdot \frac{L}{4}}{\frac{1}{12} M L^2 + m \frac{L^2}{16}} = \frac{m \cdot v \cdot L}{4 \left( \frac{1}{12} M L^2 + \frac{m L^2}{16} ight)} \] Simplify the expression: \[ \omega_f = \frac{m \cdot v \cdot L}{4 \left( \frac{M L^2}{12} + \frac{m L^2}{16} ight)} = \frac{m \cdot v}{4 \left( \frac{M}{12} + \frac{m}{16} ight)} \] Combine the terms in the denominator: \[ \omega_f = \frac{m \cdot v}{4 \left( \frac{4M + 3m}{48} ight)} = \frac{48m \cdot v}{4 (4M + 3m)} = \frac{12m \cdot v}{4M + 3m} \] ### Final Results - **Translational Velocity (v_f):** \[ v_f = \frac{m \cdot v}{M + m} \] - **Angular Velocity (\(\omega_f\)):** \[ \omega_f = \frac{12m \cdot v}{4M + 3m} \] These equations give the translational and rotational velocities of the rod after the collision. Yunus Ghausi, P.E , T.E, PPM, MS. Yunus.Ghausi@outlook.com
Hello sir My answer is V= mv/[m+M] conserving linear momentum W about CM in clockwise direction =12mv/[3mL+4ML] by conserving angular momentum about CM Thank u sir for the questions
Solutions: vf = m*v/(M + m) ωf = 12*m*v/[ℓ*(4*M + 7*m)] Bodies that are both free to translate and rotate, will rotate about their center of mass. Therefore, a convenient point for analyzing this situation, is the combined center of mass (ccm) of the rod and clay ball. Conservation of momentum and angular momentum govern. Kinetic energy will not be conserved, since it's an inelastic collision. Let y be the distance from center of the rod to combined center of mass: y = m*(ℓ/4)/(M + m) Initial momentum: p = m*v Final momentum: p = (M+m)*vf Equate and solve for vf: vf = m*v/(M + m) Initial angular momentum, about ccm: L = m*v*[ℓ/4 - y] Substitute y & simplify: L = v*ℓ*M*m/[4*(M + m)] Equate to final angular momentum about ccm: L = I_net_ccm*ωf Solve for ωf: ωf = v*ℓ*M*m/[4*(M + m)*Ɪ_net_cm] Enumerate the rotational inertia, starting with the rod's rotational inertia about its original center: Ɪ_rod0 = 1/12*M*ℓ^2 Rod rotational inertia, about the ccm. Use Steiner's parallel axis theorem, with an offset distance of y. Ɪ_rod_ccm = 1/12*M*ℓ^2 + M*y^2 Clay ball rotational inertia, about ccm: Ɪ_clay_ccm = m*(ℓ/4 - y)^2 Combine: Ɪ_net_ccm = 1/12*M*ℓ^2 + M*y^2 + m*(ℓ/4 - y)^2 Plug in y, and simplify: Ɪ_net_ccm = 1/12*M*ℓ^2 + M*[m*(ℓ/4)/(M + m)]^2 + m*[ℓ/4 - m*(ℓ/4)/(M + m)]^2 Ɪ_net_ccm = M*ℓ^2*(7*m + 4*M)/[48*(m + M)] Substitute into previously-derived formula for ωf, and simplify: ωf = 12*m*v/[ℓ*(4*M + 7*m)]
Last days in Bali. Don't know if i can make time. This looks a bit taylor made for me if i understood your personal explanation or not. One thing i notice is that the base of this rod is unknown but thin. Obviously when the base is large, the rod will not rotate. The picture and text indicate a thin base so i guess it can be assumed 0.
Hi there Mr. Walter! I used the law of conservation of angula momentum in order to solve this particular problem: Lib = L fb + L fr Where b stands for the ball and r for the rod. The angular velocity of the system turns out to be: wf = (12mbVib)/(3mb + 4Mr)L Where mb and Mr are the masses of the ball and the rod respectively, Vib is the initial velocity of the ball and L is the length of the rod The linear velocity of this system is: Vf = (12mbVib)/(3mb + 4Mr) Do tell me if I solved it correct!
Hello pls help me i want to crack the NEET my understading of physic is so poor pls tell me what to do i realy need your help you are a great teacher i attended the series 801 but cant clear the concepts pls help sir give a way to understand and lear it
you have 2 options option 1: eat yogurt every day but *never on Fridays* That worked well for Einstein and also for me option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, Solutions & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
you have 2 options option 1: eat yogurt every day but *never on Fridays* That worked well for Einstein and also for me option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, Solutions & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
Prof quick question. T of a simple pendulum is largely independent of the angle θ for small angles right? But could you explain what happens to the T for larger angles? Why the period does indeed change?
Sir I am from India and I want you to make a whole video on the solution of the JEE Advanced paper of the year 2016.I will be very greatful to you for this. Please sir, I request you to make a video on the mentioned topic.
I do not solve problems that are not my own - I teach Physics, If you take all my 3 MIT courses (8.01, 8.02 and 8.03) do all the homework and take all my exams, then you can solve the JEE advanced problems on your own.
I want you solve thise because the reason is that there in not any proper solution on the internet. And if there is so the Solutions are of half an hour to a hour.
Can you please explain how to be really good at physics because at sometimes i know things but they don't strike at the time of solving questions and later when i identify it i feel very stupid to not get it .Sir i'd be very grateful if you could help me with that.
option 1: eat yogurt every day but *never on Fridays* That worked well for Einstein and also for me option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, Solutions & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
@@lecturesbywalterlewin.they9259 😂option 2 would be great I guess and if that's you're gaurantee then it surely will help thanks alot for the advice sir .
Ohm's law is that resistance is independent of electrical operating conditions. Since temperature governs resistance, and because a resistor's temperature is indirectly determined by electrical operating conditions, it can make it seem like Ohm's law "isn't valid" for resistances that heat up significantly. This is why a 100 Watt/120V bulb when cold doesn't measure as 144 Ohms on an ohm-meter, but it is 144 Ohms in its operating state. If you control temperature to remain constant, Ohm's law will be valid, for materials that obey Ohm's law, no matter what the temperature. One way you could set up such an experiment (anticipating all safety issues and temperature limitations, of course), is with a circuit where the temperature rise due to ohmic heating is insignificant, and put it in a temperature-controlled oven or refrigerator.
Sir sorry to ask but i have a doubt regarding another topic. Regarding CIRCULAR MOTION. Sir if centripetal force acts inwards and velocity of object is translational , then why is the object pushed outwards rather than inwards?? Pls tell
a rotating frame is not an inertial reference frame, thus Newton's laws do not apply. Pseudo forces are therefore introduced in the rotating reference frame. It's called centrifugal force which is pointing outwards. A pseudo force, also called a fictitious force or an inertial force, is an apparent force that acts on all bodies whose motion is described using a non-inertial frame of reference, such as a rotating reference frame. A pseudo force arises when a frame of reference is accelerating compared to a non-accelerating frame.
@@lecturesbywalterlewin.they9259so sir centrifugal forces are just assumed to act on the object to explain its motion? (While in reality it doesn't exist)
@@sivaprasadmalkapuram2247 that is not correct. *A rotating object In my inertial reference frame experiences a centripetal force in my reference frame.* If now you move to the rotating ref frame which is not an inertial ref frame, now you have to introduce a "pseudo" force which is the centrifugal force which is outwards. *If you were also in that rotating frame you would experience that centrifugal force* - *that force does exist in the rotating ref frame*
@@lecturesbywalterlewin.they9259that means i need to add both those inertial and non inertial reference frame while applying it to a problem involving a drifting car on a banked road ?
Hello sir i am a 9 class Student And I want to crack JEE exams and I am requesting u to give me some tips so that my physics will become very strong...thanku
you have 2 options option 1: eat yogurt every day but *never on Fridays* That worked well for Einstein and also for me option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, Solutions & Lecture Notes". 8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
What about the Foucault pendulum irregular movements at Sun eclipses? Nasa scientist could not explain. Is this true? I want to know the answer from a credible person. Thank you! 👍👍
@@lecturesbywalterlewin.they9259 Thank you for the quick reply, I could not think better person than you to answer for this, this information is spreading and an average person has no way to find out if it is true or false. Thanks again. 👍👍
@@lecturesbywalterlewin.they9259 Thank you for the quick reply, I could not think better person than you to answer for this, this information is spreading and an average person has no way to find out if it is true or false. Thanks again. 👍👍
Professor, I would greatly appreciate it if you could share insights on the secrets behind the longevity often observed in Jewish communities. I look forward to your response and will keep commenting on your videos until you address this (please avoid common answers like eating yogurt).Thank you!
INAPPROPRIATE COMMENT. Professor Lewin is a physicist and that is the sole purpose of his videos. Questions like yours could easily be considered racist and inherently offensive, but I will give you the benefit of the doubt and consider it just social ignorance. If you are truly interested, look it up. PS. Maybe you would like to ask some Chinese person where to go for the best Chinese food or a black person why he thinks blacks have a disproportionate prison population.
Despite being 88 years old he is still teaching physics . That is an achievement.❤❤❤
His lecture's will be there long after passing away.
(a) Translational motion.
------------------------------
pᵢ = p₀
=> mv = (M + m)v₀
=> Translational velocity after collision:
Answer (a):
v₀ = mv/(M + m)
(b) Rotational motion.
----------------------------
Total angular momentum before collision:
Lᵢ = mvL/4
Moment of inertia of the rod before collision:
Iᵣᵢ = (1/12)ML²
Center of mass after collision:
Let y = 0 at the bottom of the rod.
Then y_cm = L/2 at the original center of mass, and
y_b = 3L/4 where the ball sticks to the rod.
The new center of mass is then located at:
y_cm’ = (M·L/2 + m·3L/4)/(M+m)
y_cm’ = (2M + 3m)/(M + m)·L/4
Moment of inertia after collision:
Let d = distance between old and new center of mass.
d = y_cm’ - y_cm
d = (2M + 3m)/(M + m)·L/4 - L/2
d = m/(M+m)·L/4
Let r = distance between new center of mass and the
point where the ball hits the rod.
r = y_b - y_cm’
d + r = L/4 => r = L/4 - d
r = M/(M+m)·L/4
Total moment of inertia after collision:
I₀ = Iᵣᵢ + Md² + mr²
where Iᵣᵢ + Md² is the moment of inertia of the rod
after collision due to the parallell axis theorem.
mr² is moment of inertia added by the ball.
I₀ = (1/12)ML² + M[m/(M+m)]²·L²/16 +
+ m[M/(M+m)]²·L²/16 … (1)
Conservation of angular momentum:
L₀ = I₀·ω₀ = Lᵢ
ω₀ = Lᵢ/I₀
Answer (b):
ω₀ = (mvL/4)/I₀
where I₀ is defined by Eqn. (1)
please sir we want to see you in physics wallah app🥺
Bruh y
Baulath
Hello sir I am from India And its 11:07 PM here
Love from india❤❤
The CM of the ball-on-rod system is (m / M+m) * L/4 from the middle of the rod. The motion of the system is the superposition of CM translating with speed (m / M+m) * v to the right and the system rotating clockwise about the CM with angular speed (12m / 4M+7m) * v/L.
Sir your lactures helping me in NEET exam too much ❤
You are the man how can never retire .
The translational motion of the pivot point after collision is to the right at speed V= v*(m/(m+M).
The pivot point after collision is NOT the center of the rod.
Here is my plan of attack for the rotation (which I have not completed at this time):
The centrifugal force of the rod portion above the pivot plus the clay mass must equal the centrifugal force of the portion of the rod below the pivot in order for there to be no vertical translation. Use a coordinate system based at the center of the rod, and call the pivot position rp. Then calculate using moments of inertia about rp:
1) Upper portion of rod above rp: integral[(wM/L)r^2]dr, over limits rp to L/2; (check - this gives the correct value for tje upper half of a rotating rod without the clay mass)
2) Clay ball: mw((L/2)-rp);
3) Lower portion of rod below rp: same form as the part above rp, except limits are -L/2 to rp.
Then use (1) + (2) - (3) = 0 to calculate rp.
Fianlly, plug rp into (2) to find w.
I hope this works!
To solve this problem, we need to consider both the conservation of linear momentum and the conservation of angular momentum, as well as the conditions of a perfectly inelastic collision (the ball sticks to the rod).
### 1. Linear Momentum Conservation
Before the collision, the linear momentum of the system is due to the clay ball:
\[ p_{\text{initial}} = m \cdot v \]
After the collision, the clay ball and the rod stick together and move with a common velocity \( v_f \):
\[ (M + m) \cdot v_f = m \cdot v \]
Solving for \( v_f \):
\[ v_f = \frac{m \cdot v}{M + m} \]
### 2. Angular Momentum Conservation
We need to calculate the angular momentum about the center of mass (CM) of the rod before and after the collision.
Before the collision, the clay ball has angular momentum about the CM of the rod:
\[ L_{\text{initial}} = m \cdot v \cdot \left( \frac{L}{4}
ight) \]
Since the rod is initially at rest, it has no initial angular momentum.
After the collision, the combined system (rod + ball) will rotate about the CM. The moment of inertia \( I \) of the rod about its center is:
\[ I_{\text{rod}} = \frac{1}{12} M L^2 \]
The moment of inertia of the clay ball about the CM of the rod (shifted by \( L/4 \)) is:
\[ I_{\text{ball}} = m \left( \frac{L}{4}
ight)^2 = m \frac{L^2}{16} \]
The total moment of inertia \( I_{\text{total}} \) after the collision is:
\[ I_{\text{total}} = \frac{1}{12} M L^2 + m \frac{L^2}{16} \]
The final angular momentum is:
\[ L_{\text{final}} = I_{\text{total}} \cdot \omega_f \]
where \( \omega_f \) is the final angular velocity. From the conservation of angular momentum:
\[ m \cdot v \cdot \frac{L}{4} = \left( \frac{1}{12} M L^2 + m \frac{L^2}{16}
ight) \cdot \omega_f \]
Solving for \( \omega_f \):
\[ \omega_f = \frac{m \cdot v \cdot \frac{L}{4}}{\frac{1}{12} M L^2 + m \frac{L^2}{16}} = \frac{m \cdot v \cdot L}{4 \left( \frac{1}{12} M L^2 + \frac{m L^2}{16}
ight)} \]
Simplify the expression:
\[ \omega_f = \frac{m \cdot v \cdot L}{4 \left( \frac{M L^2}{12} + \frac{m L^2}{16}
ight)} = \frac{m \cdot v}{4 \left( \frac{M}{12} + \frac{m}{16}
ight)} \]
Combine the terms in the denominator:
\[ \omega_f = \frac{m \cdot v}{4 \left( \frac{4M + 3m}{48}
ight)} = \frac{48m \cdot v}{4 (4M + 3m)} = \frac{12m \cdot v}{4M + 3m} \]
### Final Results
- **Translational Velocity (v_f):**
\[ v_f = \frac{m \cdot v}{M + m} \]
- **Angular Velocity (\(\omega_f\)):**
\[ \omega_f = \frac{12m \cdot v}{4M + 3m} \]
These equations give the translational and rotational velocities of the rod after the collision.
Yunus Ghausi, P.E , T.E, PPM, MS.
Yunus.Ghausi@outlook.com
Hello sir
My answer is
V= mv/[m+M] conserving linear momentum
W about CM in clockwise direction =12mv/[3mL+4ML] by conserving angular momentum about CM
Thank u sir for the questions
The ball sticks to the rod, which means that the ball's momentum is transferred to the rod.
correct
Hello sir grate explanation❤
great
a) transational speed: V=v* m/(M+m)
b) rotational speed: ω=12 *m/(4M+3m) * v/L
Solutions:
vf = m*v/(M + m)
ωf = 12*m*v/[ℓ*(4*M + 7*m)]
Bodies that are both free to translate and rotate, will rotate about their center of mass. Therefore, a convenient point for analyzing this situation, is the combined center of mass (ccm) of the rod and clay ball. Conservation of momentum and angular momentum govern. Kinetic energy will not be conserved, since it's an inelastic collision.
Let y be the distance from center of the rod to combined center of mass:
y = m*(ℓ/4)/(M + m)
Initial momentum:
p = m*v
Final momentum:
p = (M+m)*vf
Equate and solve for vf:
vf = m*v/(M + m)
Initial angular momentum, about ccm:
L = m*v*[ℓ/4 - y]
Substitute y & simplify:
L = v*ℓ*M*m/[4*(M + m)]
Equate to final angular momentum about ccm:
L = I_net_ccm*ωf
Solve for ωf:
ωf = v*ℓ*M*m/[4*(M + m)*Ɪ_net_cm]
Enumerate the rotational inertia, starting with the rod's rotational inertia about its original center:
Ɪ_rod0 = 1/12*M*ℓ^2
Rod rotational inertia, about the ccm. Use Steiner's parallel axis theorem, with an offset distance of y.
Ɪ_rod_ccm = 1/12*M*ℓ^2 + M*y^2
Clay ball rotational inertia, about ccm:
Ɪ_clay_ccm = m*(ℓ/4 - y)^2
Combine:
Ɪ_net_ccm = 1/12*M*ℓ^2 + M*y^2 + m*(ℓ/4 - y)^2
Plug in y, and simplify:
Ɪ_net_ccm = 1/12*M*ℓ^2 + M*[m*(ℓ/4)/(M + m)]^2 + m*[ℓ/4 - m*(ℓ/4)/(M + m)]^2
Ɪ_net_ccm = M*ℓ^2*(7*m + 4*M)/[48*(m + M)]
Substitute into previously-derived formula for ωf, and simplify:
ωf = 12*m*v/[ℓ*(4*M + 7*m)]
Good problem. Thanks Sir for enlightenment. Regards
Last days in Bali. Don't know if i can make time. This looks a bit taylor made for me if i understood your personal explanation or not. One thing i notice is that the base of this rod is unknown but thin. Obviously when the base is large, the rod will not rotate. The picture and text indicate a thin base so i guess it can be assumed 0.
Ans : translational motion=mv/M+m and Rotational motion = mv/4L × 1/ M/12 + m/16
Thank u sir . U are my favourite professor. Love from India ❤❤❤❤
Hi there Mr. Walter! I used the law of conservation of angula momentum in order to solve this particular problem:
Lib = L fb + L fr
Where b stands for the ball and r for the rod.
The angular velocity of the system turns out to be:
wf = (12mbVib)/(3mb + 4Mr)L
Where mb and Mr are the masses of the ball and the rod respectively, Vib is the initial velocity of the ball and L is the length of the rod
The linear velocity of this system is:
Vf = (12mbVib)/(3mb + 4Mr)
Do tell me if I solved it correct!
Great explanation sir❤️🇮🇳
Sir I'm reading in class 9th and I want to know about the Lagrange's equation....please explain it to me.....
en.wikipedia.org/wiki/Lagrangian_mechanics
@@lecturesbywalterlewin.they9259 Thank you sir 🙏
Sir what I do I can't understand all because of my 10 class studying but topics that I learned I am good
Hello pls help me i want to crack the NEET my understading of physic is so poor pls tell me what to do i realy need your help you are a great teacher i attended the series 801 but cant clear the concepts pls help sir give a way to understand and lear it
you have 2 options
option 1: eat yogurt every day but *never on Fridays* That worked well for Einstein and also for me
option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, Solutions & Lecture Notes".
8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
@@lecturesbywalterlewin.they9259 thank you sir for this valueable suggestion...
Sir I am in grade 9. I want to crack JEE exam. Sir please give me some tips 🙏.
you have 2 options
option 1: eat yogurt every day but *never on Fridays* That worked well for Einstein and also for me
option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, Solutions & Lecture Notes".
8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
@@lecturesbywalterlewin.they9259 Thank you so much sir ❤️. May God bless you.
Well wishes, sir.
Same to you!
Hello professor, do you live alone in the forest?
In your opinion, which branch of physics is the most beautiful?
beauty is in the mind of the beholder
Prof quick question.
T of a simple pendulum is largely independent of the angle θ for small angles right? But could you explain what happens to the T for larger angles? Why the period does indeed change?
ruclips.net/video/1TR856BWAX0/видео.html
@@lecturesbywalterlewin.they9259 thank you prof ❤️
I am from India and I am preparing for jee please guide me
Sir could you give suggestions for how to prepare for mathematics and chemistry of JEE
I cannot
Sir I am from India and I want you to make a whole video on the solution of the JEE Advanced paper of the year 2016.I will be very greatful to you for this. Please sir, I request you to make a video on the mentioned topic.
I do not solve problems that are not my own - I teach Physics, If you take all my 3 MIT courses (8.01, 8.02 and 8.03) do all the homework and take all my exams, then you can solve the JEE advanced problems on your own.
Please sir. One time only I want you to make a solution video on the physics section of JEE Advanced 2016.Please sir please.
I want you solve thise because the reason is that there in not any proper solution on the internet. And if there is so the Solutions are of half an hour to a hour.
The questions were so tough that even the teachers of big institutions can't solve it.
Can you please explain how to be really good at physics because at sometimes i know things but they don't strike at the time of solving questions and later when i identify it i feel very stupid to not get it .Sir i'd be very grateful if you could help me with that.
option 1: eat yogurt every day but *never on Fridays* That worked well for Einstein and also for me
option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, Solutions & Lecture Notes".
8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
Sir when I completed 8.01 8.02 8.03 what is the level of my physics syllables
Or degree
@@user-su9cl8pw2ephysics undergraduate 2nd year I think bcoz 8.01 and 8.02 are mostly taught in 1st year and 8.03 in 2nd year 🎉
@@lecturesbywalterlewin.they9259 😂option 2 would be great I guess and if that's you're gaurantee then it surely will help thanks alot for the advice sir .
Sir I have a question about current electricity...Is it true that Ohms law valid only under lower temperatures...Please explain this to me sir...😢🙏🙏🙏
I cover this in my 8.02 lectures - watch them, that's what they are for.
you can also use google
@@lecturesbywalterlewin.they9259 OK thank you sir...I will watch...🥰🙏
@@lecturesbywalterlewin.they9259 Ok sir...I will...🙏🙏🥰
Ohm's law is that resistance is independent of electrical operating conditions. Since temperature governs resistance, and because a resistor's temperature is indirectly determined by electrical operating conditions, it can make it seem like Ohm's law "isn't valid" for resistances that heat up significantly. This is why a 100 Watt/120V bulb when cold doesn't measure as 144 Ohms on an ohm-meter, but it is 144 Ohms in its operating state.
If you control temperature to remain constant, Ohm's law will be valid, for materials that obey Ohm's law, no matter what the temperature. One way you could set up such an experiment (anticipating all safety issues and temperature limitations, of course), is with a circuit where the temperature rise due to ohmic heating is insignificant, and put it in a temperature-controlled oven or refrigerator.
Sir sorry to ask but i have a doubt regarding another topic.
Regarding CIRCULAR MOTION.
Sir if centripetal force acts inwards and velocity of object is translational , then why is the object pushed outwards rather than inwards?? Pls tell
a rotating frame is not an inertial reference frame, thus Newton's laws do not apply. Pseudo forces are therefore introduced in the rotating reference frame. It's called centrifugal force which is pointing outwards.
A pseudo force, also called a fictitious force or an inertial force, is an apparent force that acts on all bodies whose motion is described using a non-inertial frame of reference, such as a rotating reference frame. A pseudo force arises when a frame of reference is accelerating compared to a non-accelerating frame.
@@lecturesbywalterlewin.they9259so sir centrifugal forces are just assumed to act on the object to explain its motion? (While in reality it doesn't exist)
@@sivaprasadmalkapuram2247 that is not correct. *A rotating object In my inertial reference frame experiences a centripetal force in my reference frame.* If now you move to the rotating ref frame which is not an inertial ref frame, now you have to introduce a "pseudo" force which is the centrifugal force which is outwards. *If you were also in that rotating frame you would experience that centrifugal force* - *that force does exist in the rotating ref frame*
@@lecturesbywalterlewin.they9259that means i need to add both those inertial and non inertial reference frame while applying it to a problem involving a drifting car on a banked road ?
This seems familiar sir.
yyayayaya first one to my favourite teacher's new video
♥️
Hello
Hello sir i am a 9 class Student And I want to crack JEE exams and I am requesting u to give me some tips so that my physics will become very strong...thanku
you have 2 options
option 1: eat yogurt every day but *never on Fridays* That worked well for Einstein and also for me
option 2: Watch all my 94 MIT course lectures. Start with 8.01, then 8.02, then 8.03. Do all the homework and take all my exams. *I guarantee you that you will then do very well on the Physics portion of any freshman college or JEE exam* You will find all information you need on this channel in three playlists "Homework, Exam, Solutions & Lecture Notes".
8.01 & 8.02 will each take about 200 hours, 8.03 about 250 hours.
I'm Indian 🤝
Sir I am not able to concentrate in class I don't know what should I do
eat yoghurt every day but never on Fridays - that also worked for Einstein and for me
@@lecturesbywalterlewin.they9259 Sir please tell us the secret, we deserve to live a long life!
What about the Foucault pendulum irregular movements at Sun eclipses? Nasa scientist could not explain. Is this true? I want to know the answer from a credible person. Thank you! 👍👍
this is utter nonsense
this is utter nonsense
@@lecturesbywalterlewin.they9259 Thank you for the quick reply, I could not think better person than you to answer for this, this information is spreading and an average person has no way to find out if it is true or false. Thanks again. 👍👍
@@lecturesbywalterlewin.they9259 Thank you for the quick reply, I could not think better person than you to answer for this, this information is spreading and an average person has no way to find out if it is true or false. Thanks again. 👍👍
I am from India and you
There is only a translational motion: v/v'=(m + M)/m
First comment please sir like I will be happy 🥹🥹❤️❤️✨ I love physics
Hi sir
First comment from ecuador
Professor, I would greatly appreciate it if you could share insights on the secrets behind the longevity often observed in Jewish communities. I look forward to your response and will keep commenting on your videos until you address this (please avoid common answers like eating yogurt).Thank you!
INAPPROPRIATE COMMENT. Professor Lewin is a physicist and that is the sole purpose of his videos. Questions like yours could easily be considered racist and inherently offensive, but I will give you the benefit of the doubt and consider it just social ignorance. If you are truly interested, look it up.
PS. Maybe you would like to ask some Chinese person where to go for the best Chinese food or a black person why he thinks blacks have a disproportionate prison population.
First one❤