A copper block rests 30 cm from center of a turntable. Coefficient of static fri

Question

A copper block rests 30 cm from center of a turntable. Coefficient of static friction between surface and block is 0.53. Turntable starts from rest and rotates at a constant angular acceleration of 0.5 rad/s^2. At what time wil block start to slip? (hint: normal force= weight of block) I found some help with this on yahoo. The set up makes sense as I went through it, but my answer was wrong. I used 3x10-2m for the radius. Not certain if this wasn't the right way to do it, or if my math was wrong. Answer is 8.3 s.

u = coefficient of static friction

u*m*g=m*a

cancel the masses, and substitute (V^2)/R for a

u*g=(V^2)/R

Now, recall that Vfinal=At

u*g=(a^2)*(t^2)/R

Now, solve for t and plug in the variables.

sqrt[u*g*R/(a^2)]=(t^2)

thank you!

Explanation / Answer

Let after time t is starts slipping.
angular velocity of the block = t = .5t

v = r

so force acting on the block after time t = m^*r = .25 m t^2 r

frictional force = mg

so

mg = .25 m t^2 r

on solving for t we get

t = 8.32 sec

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