A string is wrapped around a uniform solid cylinder of radius , as shown in the

Question

A string is wrapped around a uniform solid cylinder of radius , as shown in the figure . The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass . Note that the positive y direction is downward and counterclockwise torques are positive Find the magnitude of the angular acceleration of the cylinder as the block descends. Express your answer in terms of the cylinder's radius and the magnitude of the acceleration due to gravity .

Explanation / Answer

The sum of the forces acting at the block provide us this eqaution

mbg - T = mba (Since the weight of the block acts down and the tension in the string acts up)

The tension also provides a torque to spin the cylinder

= Tr (Tension times radius)

also = I, so Tr = I

I for a cylinder is .5mcr2    and = a/r, so the equation becomes

Tr = .5mcr2a/r

Simplify

T = .5mca

Substitute back into the first equation

mbg - .5mca = mba

mbg = mba + .5mca

mbg = a(mb + .5mc)

a = mbg/(mb + .5mc)

and = a/r so

= mbg/r(mb + .5mc)

Now, the answer wants the acceleration in terms of the acceleration due to gravity and the radius. However, to eliminate the masses, I believe you may have typed the question with an incomplete statement. You said that the block and the cylinder each have mass. If you meant they had EQUAL mass, the masses cancel and you get a different answer than the above. If not, the answer above is as far as it can be taken for .

Here is the answer if you meant the block and cylinder had equal masses.

a = 2/3(g/r) or .667g/r or g /1.5r   (all the same, just different ways to state the answer)

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