# When two objects roll down an incline, does the velocity increase?

I know the basics of rotational motion but this question just confused everything: The answer to the question is A. But why?

My problems:

1. If first, I treated both the disks as two particles, they would both have to accelerate down the incline at $gsin\theta$. Thus, going back to the rigid object, the tangential velocity can't remain constant at the first place

2. If both the objects started from rest and ended at the end of the incline at the same time, shouldn't they have the same tangential acceleration and thus answer should be B?

• If you think the answer is B, then try and figure out why the other options are all wrong. Sep 29 '14 at 13:34
• @Jasper: I could eliminate the last 3 options easily. The first two are causing problems... Sep 29 '14 at 13:44
• Suppose their radii are 1cm and 2cm. Since it says they roll together, they have to have the same tangential velocity (i.e. zero at the point of contact, and 2V at the opposite point. The "axle system" must be one that turns the small one at twice the angular velocity (with gears). Since centripetal acceleration if V/R^2, the small one has 4 times the acceleration. Sep 29 '14 at 13:57
• @ Mike Dunlavey: I am still confused with the second point, the question is referring to $angular$ acceleration and not $centripetal$ acceleration in B Sep 29 '14 at 14:08
• Who thinks up these terrible questions? The question doesn't say whether the axle/wheel connection is rigid or involves bearings, doesn't say how the wheels are constrained to move together (presumably with a track), and doesn't say whether the wheels are rolling with or without slipping. One thing is certain: If the tangential velocities are the same, then so are the tangential accelerations. Regarding C, D, and E, whether those are also true depends on how one interprets the question. Sep 29 '14 at 16:21

• :I see, so at every $instant$, the tangential velocity is the same for both disks even if the tangential velocity DOES increase as the disk rolls down. However, I still can't convince myself why the angular acceleration is not the same... Sep 29 '14 at 14:06
• To understand the angular acceleration, it should be clear if you take the ratio of the relation $v=\omega R$ for the two discs. Sep 29 '14 at 14:11
• I think that you are correct. Both the tangential velocity and the acceleration are the same at each instant. At the point of contact tangential velocity is zero: $-\omega R+ |V|=0$ with $V$ the velocity down the incline. Thus $|V|=\omega_1 R_1=\omega_2 R_2$. At the opposite side of the discs the velocity is $\omega R+|V|=2|V|$. In general the tangential velocity is some function $v_t(|V|,\phi)$ with $\phi$ the angle on the disc. The tangential acceleration is $dv_t/dt=(dv_t/d|V|) (d|V|/dt)$. It only depends on how the velocity down the incline changes. Sep 29 '14 at 16:44