The two blocks in (Figure 1) are connected by a massless rope that passes over a

Question

The two blocks in (Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 13cm in diameter and has a mass of 2.0kg . As the pulley turns, friction at the axle exerts a torque of magnitude 0.60N?m

If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

he two blocks in (Figure 1) are connected by a massless rope that passes over a pulley. The pulley is 13cm in diameter and has a mass of 2.0kg . As the pulley turns, friction at the axle exerts a torque of magnitude 0.60N?m If the blocks are released from rest, how long does it take the 4.0 kg block to reach the floor?

Explanation / Answer

Sum forces about each block and torques about the pulley to find the acceleration

Then use kinematic eqn to find time

So m1*g - T1 = m1*a or T1 = m1*g - m1*a

for the 2kg block we have T2 - m2*g = m2*a or T2 = m2*g + m2*a

for the pulley we have T1*r - 0.70 - T2*f = I*? where I = 1/2*M*r^2 and ? = a/r

so we have (m1*g - m1*a)*r - 0.70 -(m2*g + m2*a) *r = 1/2*M*r^2*a/r = 1/2*M*r*a

rearranging terms

a*(1/2*M*r + m1*r + m2*r) = m1*g*r - m2*g*r -0.60

or a*(1/2*2*0.065 + 4.0*0.065 + 2.0*0.065) = 4*9.8*0.065 - 2.0*9.8*0.065 - 0.60

so a = 0.674/0.455 = 1.481m/s^2

So the 4 kg block must fall 1.0m from rest under an acceleration of 1.481m/s^2

y = 1/2*a*t^2

or t = sqr(2*y/a) = sqrt(2*1.0/1.481) = 1.72s

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