A 3.0 kg block is attached to a string that is wrapped around a 2.0kg , 4.0 cm d

Question

A 3.0 kg block is attached to a string that is wrapped around a 2.0kg , 4.0 cm diameter hollow cylinder that is free to rotate. (Use Figure 12.34 in the textbook but treat the cylinder as hollow.) The block is released 1.0 above the ground.

What this question means is that the block is attached to a string which is wrapped around the hollow cylinder in the same way that a fishing line is wrapped around the spool on the rod.


Can you please determine the velocity of the block as it hits the ground using bothe Newton's second law and conservation of energy?

Explanation / Answer

The gravity of this M kg block is Mg, which causes the accelaration of itself downward and causes the rotational accelaration of the cylinder (mass m): Mg = Ma + ((mr^2)*a/r)/r = (M+m)a thus a = Mg/(M+m) Since the distance of the movement is 1m, we have; 0.5*at^2 = 1, or v^2 = (at)^2 = 2a = 2Mg/(M+m) v = sqrt(2*3.0*9.8/5.0) = 3.4 (m/s) Now use the energy conservation to solve this problem. the initial potential energy: Mgh = Mg The final kinetic energy: 0.5Mv^2 + 0.5I*w^2 = 0.5Mv^2 + 0.5(m*r^2)*(v/r)^2 = 0.5*(M+m)*v^2 Before M hits the ground, all the potential energy would be converted to kinetic energy: Mg = 0.5*(M+m)*v^2 or v = sqrt(2*M*g/(M+m)) = 3.4 (m/s)

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