A thin string is wrapped several times around a pulley and is then attached to a block of mass 3.00 kg on one side The pulley is a uniform disk of mass 15.0 kg and diameter 0.700 m, and turns on a frictionless pivot. The block is released from rest, falls a height h, and hits the floor. Just before the block hits, it is moving at 2.6 m/s. What is the height h?
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The initial energy of the system is mgh and the final energy of the system is 0.5mv² + 0.5Iω². setting these equal, we have:
mgh = 0.5mv²+ 0.5Iω²
Now, the moment of inertia (I) for a uniform solid disk is 0.5MR² and angular velocity (ω) is equal to v/R, so the equation above becomes:
mgh = 0.5mv² + 0.5(0.5MR²)(v/R)²
4mgh = 2mv² + Mv²
h = v²(2m + M) / 4mg
= (2.6m/s)²[2(3.00kg) + 15.0kg] / 4(3.00kg)(9.81m/s²)
h = 1.21m
Hope this helps.
mgh = 0.5mv²+ 0.5Iω²
Now, the moment of inertia (I) for a uniform solid disk is 0.5MR² and angular velocity (ω) is equal to v/R, so the equation above becomes:
mgh = 0.5mv² + 0.5(0.5MR²)(v/R)²
4mgh = 2mv² + Mv²
h = v²(2m + M) / 4mg
= (2.6m/s)²[2(3.00kg) + 15.0kg] / 4(3.00kg)(9.81m/s²)
h = 1.21m
Hope this helps.