Recent content by mancity

E10-8 is attached. I think where the "discrepancy" lied for my first approach was that I considered delta L_lin=Rm(v-v_0), but I thought that v was the translational velocity (incorrect), not the tangential one. Delta L_rot=I(w-w_0) is a rotational/tangential velocity. When writing...

Thanks. The context for my question is this problem, and its corresponding solution: My logic was: here, the ball rotates in the counterclockwise direction, while the translational velocity of the ball is to the left (such that the point of contact of the ball with the ground rolls without...

In this derivation above, I have to account for the fact that v is translational and opposite to the sign of w, and similarly for v_0 and w_0, so the real equation should look something like this: Now, what I don't understand is, in this second derivation using net change in linear and...

Thank you. To clarify: prior to when the object starts its rolling without slipping, the axis of rotation is the CM of the ball, but after rolling without slipping begins, the axis of rotation is (instantaneously) the point of contact between the ball and the floor?

The solution is also attached. I understand everything here except for part (b). What I don't get - why is it that when the ball starts rolling without slipping, we use moment of inertia about the point of contact instead of about the center of mass; and why is it that when the ball starts...

I apologize if this is a stupid question but how come we can't just say that for part (a) v_A = -v_1 j hat, and for (b) v_A=v2 sintheta i hat + (-v1-v2cos theta) j hat? i.e. how come we can't just do vector decomposition "normally"? I am especially confused about the given solution in (a) where...

In the. solution attached I'm not too sure why in frame K, we apply the doppler effect twice. Also, since the photon is moving away from the source, shouldn't the signs be switched? Thanks

Sorry. I guess my confusion is that a massless pulley with a mass m hanging off of it does not make the pulley massive/ have a moment of inertia?

Say you had a pulley with a mass hanging off of it, like in the picture. What I don't understand is, what idealizations are being made such that there is no net torque being generated? My confusion is that we have a massive pulley, hence there will be rotational inertia. But it is pretty clear...

I'm not quite sure how to apply conservation of string to this problem, so guidance would be appreciated. Normally as long as there isn't a "sub-pulley" I can do the problem fairly easily but this one tricks me up. Thanks

Doesn't friction always oppose the motion? From the clockwise rotation here, shouldn't the cylinder be moving to the right? so why are the acceleration and friction in the same direction to the right, and in the same direction as the motion? (attached image for reference)

Albeit the simple question, I am a bit confused on whether the correct answer choice is (B) or (C). When the piano is slowing down, shouldn't the force received by the piano be a bit greater than the force received by the man?

The correct solution, as given by the system, is 1.8s.

The correct answer as given by the system is III & IV. I believe small angles were indeed the intent for this problem.

So, per my understanding, would the correct answer then be III & IV? (because that is indeed an answer choice. all other answer selections only have one, i.e. I, or II, or III, or IV). Because I believe III only holds for small angles (in which we can approximate theta=sin(theta)), which would...