please answeer both! A sledgehammer with a mass of 2.10kg is connected to a fric

Question

please answeer both!

A sledgehammer with a mass of 2.10kg is connected to a frictionless pivot at the tip of its handle. The distance from the pivot to the center of mass is r_cm=0.550m, and the moment of inertia about the center of mass is I_cm = 0.0380 kg*m2. If the hammer is released from rest at an angle of theta=45.0 degrees such that H=0.389m, what is the speed of the center of mass when it passes through horizontal? A ball of mass 2.50kg and radius 0.144m is released from rest on a plane inclined at an angle theta=41.0 degrees with respect to the horizontal. How fast is the ball moving (in m/s) after it has rolled a distance d=1.40m? Assume that the ball rolls without slipping, and that its moment of inertia about its center of mass is 1.30 10 - 2 kg*m2.

Explanation / Answer

Ipivot = I_cm + M*R^2

= 0.0380 + 2.10 * 0.550^2

= 0.673 kg?m2

Conservation of mechanical energy yields

m*g*H = 1/2*m*v^2 + 1/2*I*?^2 = 1/2*m*v^2 + 1/2*I*(v/r)^2

m*g*H = (1/2*m + 1/2*I/r^2)*v^2

(1/2 * 2.10 + 1/2 * 0.673 / 0.550^2) * v^2 = 2.1 * 9.8 * 0.389

v = 1.92 m/s

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By work - energy theorem ,

m g d sin theta = 1/2 mv^2 + 1/2 Iw^2

(2.5 g)(1.40)sin41 = 1/2 (2.5)v^2 + 1/2 (1.3 X 10-2 ) w^2

45 = 2.5v^2 + 0.013(v/r)^2 (as v =wr , so w = v/r)

45 = 2.5 v^2 + 0.627v^2

v = 3.99 m/s

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