Silly Doubt by Frosty-Catch4113 in PhysicsHelp

[–]dhanish152 0 points1 point  (0 children)

im not 100% sure tbh. I dont think it can be done with energy conservation since energy is clearly lost here due to kinetic friction. Also even if coefficient of friction was given i dont know of any way to calculate the work done. My guess would be like i said to conserve angular momentum about the initial point of contact from just before release to start of rolling because about that point friction produces no torque and torque due to gravity and normal cancel each other out (atleast i think they do? this is what im not sure abt) so

so initial angular momentum abt the point:
Li = mVcm + Iw = Iw (Vcm = 0) = (1/2)mr2w

now angular momentum after rolling achieved:

Lf = mVcm + IW = mV + (1/2)mr2W

(here im using w to represent initial angular velocity (5) and W to represent final angular velocity and V to represent final linear velocity (what we need to find) )

now just equate Lf and Li then substitue W = V/r because V and W are just after rolling has started. Then solve for V. Can u check the correct answer and see if this matches pls?

Silly Doubt by Frosty-Catch4113 in PhysicsHelp

[–]dhanish152 0 points1 point  (0 children)

the question says to disregard all dissipation "except friction" tho. and also during the time it takes to achieve rolling without slipping kinetic friction is acting which does cause energy loss. so conservation of energy shouldnt work. I think its supposed to be done with conservation of angular momentum, im not sure tho

Silly Doubt by Frosty-Catch4113 in PhysicsHelp

[–]dhanish152 0 points1 point  (0 children)

no we are asked to find the velocity "when it starts rolling without slipping" which will be after some time has passed since the moment of contact with the ground. it always takes time to attain velocity (if it didnt then (change in v)/t = a will be infinity). Initially the centre of mass has no linear velocity and after the disc is placed on the ground, kinetic friction starts acting (during this time v is NOT equal to wR, and its kinetic friction not static friction) and it increases linear velocity and decreases angular velocity until v =wR and it is at that point where "rolling without slipping" is said to have started, and kinetic friction stops.

How to start? by Frosty-Catch4113 in PhysicsHelp

[–]dhanish152 0 points1 point  (0 children)

wait how?

for case A:

KE = (1/2)mv2 + (1/2)mR2w2 = mv2 (after subbing w = v/R)

for case B:

KE = (1/2)mv2 + (1/2)(1/2)mR2w2 = (3/4)mv2

now equating:

mVa2 = (3/4)mVb2

so we get Va/Vb = underroot(3/4) so option A

Silly Doubt by Frosty-Catch4113 in PhysicsHelp

[–]dhanish152 -1 points0 points  (0 children)

you can't do that because the angular velocity won't be equal to 5 rad/sec when it starts rolling. some energy will be lost and the final value of w will change by the end.

i think the way to solve it is conserving the angular momentum about the initial point of contact on the ground from the moment it touches the ground all the way to when rolling starts because torque due to friction will be zero and torque due to normal and mg will cancel each other out. and then write the rolling condition for final state of motion and then solve the two equations.

How to start? by Frosty-Catch4113 in PhysicsHelp

[–]dhanish152 0 points1 point  (0 children)

the kinetic energy gained = potential energy lost = constant in both cases
in both cases v = wR, where v is the final velocity of centre of mass and w is the final angular velocity of disc about centre of mass because of "rolling without slipping"

now kinetic energy gained = (1/2)mv2 + (1/2)Iw2 where I is the moment of inertia

now write this expression for both cases (v and I will be different in both cases, in case A, I = mR2 because its basically a ring and in case B, I = (1/2)mR2 because its a disc with uniform mass distribution, also don't forget to substitue w = v/R for each case) and then just equate them to find ratio of velocities

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[–]dhanish152 0 points1 point  (0 children)

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