Rotational Motion of a disc of mass

AI Thread Summary
A disc of mass M and radius R is rolling down an incline while dragging an additional mass M, raising questions about the linear acceleration of the system. The frictional force, denoted as fs, is crucial for determining this acceleration, but its exact value is unclear to the user. The teacher clarifies that fs equals 1/2Ma, which leads to the equation a = 2fs/M. The confusion arises from needing to find the frictional force before determining acceleration, but it is established that the acceleration must be calculated first to find the static friction. Ultimately, static friction is a passive force that adjusts based on the system's acceleration.
PhysicsNewb
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I'm stuck on a miniquestion in my HW that is hindering the entire problem, so here it is.

A disc of mass M, radius R, Icm =1/2MR^2 is rolling down an incline dragging a mass M attached with a light rod to a bearing at the center of the disc. The friction coefficients are the same for both masses, us and uk.

Determine the linear acceleration of the mass M.

Well, this question seems easy, and I ended up with 2fs/M = a.
But I need to find out what fs is to simplify my answer, and I can't figure out how.
 
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I don't see how anyone can tell you what fs IS if you don't tell us what it MEANS. I might guess, since it is "f", that it is a force and then guess that "s" means directed along the incline, but I'm still not sure whether it is the force of gravity along the incline or the retarding force due to friction.

If it is gravitational force, then it is -2Mgsin(θ) where θ is the angle the incline makes with the horizontal. (Notice the 2. Gravity pulls on both masses.)

If it is the friction force then it is μMgcos(θ) where Mgcos(θ) is the force normal to the plane due to the weight of the second mass. The disk is rolling so there is no sliding friction.
 
I'm sorry I should have been clearer. fs is the frictional force, except on my paper my teacher said that fs does not equal uMGcos(theta), which is the source of my problem. He says fs equals 1/2Ma, which is where I got a = 2fs/M from.

Sum of the torques = I(angular acceleration)
fsR = 1/2MR^2(a/R)
fs = 1/2Ma
 
Last edited:
PhysicsNewb said:
He says fs equals 1/2Ma, which is where I got a = 2fs/M from.
That comes from applying Newton's 2nd law to the rotational motion of the cylinder. (\tau = I \alpha) Now apply Newton's 2nd law to the translational motion of the entire system.
 
Okay, here is the source of my confusion about this problem. First it says find the linear acceleration of mass M, then the following question asks me to find the frictional force acting on the disc. But don't I need to find the frictional force before I find the acceleration?
 
PhysicsNewb said:
But don't I need to find the frictional force before I find the acceleration?
No. In fact, the only way you can find the static friction on the disc is to find the acceleration, since it depends on the acceleration. Try it!

(Static friction is a "passive" force--it can adjust to be whatever it needs to be up to a maximum value.)
 
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