Rolling Objects by Walter Lewin

In this video Professor Walter Lewin compares the roll motion of different objects rolling down a slope.
00:00 - Two solid cylinders of different material
00:58 - Results and discussion
01:29 - Two solid cylinders with different lengths
01:58 - Results and discussion
02:26 - Two solid cylinders with different material and radius
03:32 - Results and discussion
04:10 - One solid cylinder and one hollow cylinder
05:05 - Results and discussion
06:09 - Two solid cylinders with great difference in weight
06:39 - Results and discussion
06:53 - “No, it’s not simple” - Great Explanation by Prof. Walter Lewin

[@ProfessorPauloFlores]

for a rolling object, with mass m, descending a ramp with inclination teta, the linear acceleration is given by the following expression:
a = (m*g*sin(teta)) / (m+(I/r^2))
where I denotes the rotational inertia or mass moment of inertia at the center of mass, and r is the radius
below are the mass moment of inertia
for several rolling objects:
1. Solid Sphere: 2/5 *m*r^2
2. Hollow Sphere: 2/3 *m*r^2
3. Solid Cylinder: 1/2 *m*r^2
4. Hollow Cylinder: m*r^2
thus, corresponding accelerations are:
1. Solid Sphere: a = 5/7*g*sin(teta)
2. Hollow Sphere: a = 3/5*g*sin(teta)
3. Solid Cylinder: a = 2/3*g*sin(teta)
4. Hollow Cylinder: a = 1/2*g*sin(teta)
hence, in a race between these 4 rolling objects down in an inclined plane the results are:
1st place: Solid Sphere
2nd place: Solid Cylinder
3rd place: Hollow Sphere
4th place: Hollow Cylinder
See demonstrative video of this race in the following link (instant of time 55min 05sec):
ruclips.net/video/eXfwodnO6lc/видео.html

[@JunFetao]

Nice and complete description. Thanks 🙂

[@ProfessorPauloFlores]

@@JunFetao Many thanks.

[@JinZin-v8o]

Great lesson from Dr. Lewin. Great mentor and physician. Big Thanks.

[@ProfessorPauloFlores]

I thank you so much.

[@ProfessorPauloFlores]

for a rolling object, with mass m, descending a ramp with inclination teta, the linear acceleration is given by the following expression:
a = (m*g*sin(teta)) / (m+(I/r^2))
where I denotes the rotational inertia or mass moment of inertia at the center of mass, and r is the radius
below are the mass moment of inertia for several rolling objects:
1. Solid Sphere: 2/5 *m*r^2
2. Hollow Sphere: 2/3 *m*r^2
3. Solid Cylinder: 1/2 *m*r^2
4. Hollow Cylinder: m*r^2
thus, corresponding accelerations are:
1. Solid Sphere: a = 5/7*g*sin(teta)
2. Hollow Sphere: a = 3/5*g*sin(teta)
3. Solid Cylinder: a = 2/3*g*sin(teta)
4. Hollow Cylinder: a = 1/2*g*sin(teta)
hence, in a race between these 4 rolling objects down in an inclined plane the results are:
1st place: Solid Sphere
2nd place: Solid Cylinder
3rd place: Hollow Sphere
4th place: Hollow Cylinder
See demonstrative video of this race in the following link (instant of time 55min 05sec):
ruclips.net/video/eXfwodnO6lc/видео.html

[@petrisorcatana5735]

a= (2/3)g •sin( alfa) ,cilinder

[@ProfessorPauloFlores]

Many thanks.
In the link below it is a video with an example of application for spheres, cylinder, and ring:
ruclips.net/video/FDvcIuNEgo0/видео.html

[@ProfessorPauloFlores]

for a rolling object, with mass m, descending a ramp with inclination teta, the linear acceleration is given by the following expression:
a = (m*g*sin(teta)) / (m+(I/r^2))
where I denotes the rotational inertia or mass moment of inertia at the center of mass, and r is the radius
below are the mass moment of inertia for several rolling objects:
1. Solid Sphere: 2/5 *m*r^2
2. Hollow Sphere: 2/3 *m*r^2
3. Solid Cylinder: 1/2 *m*r^2
4. Hollow Cylinder: m*r^2
thus, corresponding accelerations are:
1. Solid Sphere: a = 5/7*g*sin(teta)
2. Hollow Sphere: a = 3/5*g*sin(teta)
3. Solid Cylinder: a = 2/3*g*sin(teta)
4. Hollow Cylinder: a = 1/2*g*sin(teta)
hence, in a race between these 4 rolling objects down in an inclined plane the results are:
1st place: Solid Sphere
2nd place: Solid Cylinder
3rd place: Hollow Sphere
4th place: Hollow Cylinder
See demonstrative video of this race in the following link (instant of time 55min 05sec):
ruclips.net/video/eXfwodnO6lc/видео.html

[@exlife9446]

this is a very important experiment, and it tells the diameter does not affect the acceleration of the cylinder rolling down on the slope.

[@ProfessorPauloFlores]

Many thanks.

[@ProfessorPauloFlores]

The acceleration can be expressed as follows: a = (2/3) g sin (teta)

[@ProfessorPauloFlores]

for a rolling object, with mass m, descending a ramp with inclination teta, the linear acceleration is given by the following expression:
a = (m*g*sin(teta)) / (m+(I/r^2))
where I denotes the rotational inertia or mass moment of inertia at the center of mass, and r is the radius
below are the mass moment of inertia for several rolling objects:
1. Solid Sphere: 2/5 *m*r^2
2. Hollow Sphere: 2/3 *m*r^2
3. Solid Cylinder: 1/2 *m*r^2
4. Hollow Cylinder: m*r^2
thus, corresponding accelerations are:
1. Solid Sphere: a = 5/7*g*sin(teta)
2. Hollow Sphere: a = 3/5*g*sin(teta)
3. Solid Cylinder: a = 2/3*g*sin(teta)
4. Hollow Cylinder: a = 1/2*g*sin(teta)
hence, in a race between these 4 rolling objects down in an inclined plane the results are:
1st place: Solid Sphere
2nd place: Solid Cylinder
3rd place: Hollow Sphere
4th place: Hollow Cylinder
See demonstrative video of this race in the following link (instant of time 55min 05sec):
ruclips.net/video/eXfwodnO6lc/видео.html

[@exlife9446]

@@ProfessorPauloFlores I appreciate for your detailed and informative complements, Dear professor.

[@ProfessorPauloFlores]

@@exlife9446 You are welcome.