Rolling Stuff Down an Incline
Most recent answer: 10/22/2007
Q:
A sphere, a solid cylinder and a thin-walled hollow cylinder all have the same mass and radius. The three are initially held side by side on an incline. They are all released simultaneously and roll down the incline. In what order will these three items reach the bottom of the incline?
I know that I can solve for acceleration for each of these items and thus determine their order down the incline. My question is, if I compare the objects moment of inertias, will that also give me an accurate account of the order down the incline plane?
Thanks
- Phil (age old)
Monticello High School, Monticello, IL
- Phil (age old)
Monticello High School, Monticello, IL
A:
Yup, comparing the moments of inertia should give you the correct
ordering, with the given conditions that the masses and radii of the
round objects are the same.
The full solution, assuming the objects do not slip as they roll down the inclined plane, is:
acceleration = g*sin(theta)/(1+I/(M*R^2))
where g is the acceleration due to gravity, 9.81 m/s^2, theta is the angle the incline makes with respect to the horizontal, I is the moment of inertia of the rolling object, M is the mass of the rolling object, and R is the radius of the rolling object.
For a uniform solid sphere, the moment of inertia I=(2/5)*M*R^2, for a hollow tube, I=M*R^2, and for a uniform solid cylinder I=(1/2)*M*R^2.
The interesting feature is that the acceleration of the object as it rolls down the incline depends on what kind of object it is and not how big or heavy it is.
You can solve for that acceleration using Newton’s second law, and also that torque=I*angular acceleration (there’s a friction force which needs to be solved for).
Tom
Another way to look at exactly the same result is to think of what happens to the energy that the thing picks up as it rolls down. Some of it goes into the net average motion and some into the turning. The more that goes into turning, the less that’s left to speed the object down the slope. That’s why the bigger "I" gives lower acceleration. /Mike W.
The full solution, assuming the objects do not slip as they roll down the inclined plane, is:
acceleration = g*sin(theta)/(1+I/(M*R^2))
where g is the acceleration due to gravity, 9.81 m/s^2, theta is the angle the incline makes with respect to the horizontal, I is the moment of inertia of the rolling object, M is the mass of the rolling object, and R is the radius of the rolling object.
For a uniform solid sphere, the moment of inertia I=(2/5)*M*R^2, for a hollow tube, I=M*R^2, and for a uniform solid cylinder I=(1/2)*M*R^2.
The interesting feature is that the acceleration of the object as it rolls down the incline depends on what kind of object it is and not how big or heavy it is.
You can solve for that acceleration using Newton’s second law, and also that torque=I*angular acceleration (there’s a friction force which needs to be solved for).
Tom
Another way to look at exactly the same result is to think of what happens to the energy that the thing picks up as it rolls down. Some of it goes into the net average motion and some into the turning. The more that goes into turning, the less that’s left to speed the object down the slope. That’s why the bigger "I" gives lower acceleration. /Mike W.
(published on 10/22/2007)