Consider a thin 24 m rod pivoted at one end. A uniform density spherical object

Question

Consider a thin 24 m rod pivoted at one end. A uniform density spherical object (whose mass is 9 kg and radius is 4.9 m) is attached to the free end of the rod and the moment of inertia of the rod about an end is I(rod) =(1/3)m L^2 and the moment of inertia of the sphere about its center of mass is I(sphere) =(2/5)m r^2. What is the angular acceleration of the rod immediately after it is released from its initial position of 41 degrees from the vertical? The acceleration of gravity g = 9.8 m/s^2. Answer in units of rad/s^2 My work= Torgue(net) = I a, a = I / (F*g*r*sin @), a ={ (2/5 * M(sphere)*r(sphere)^2 + mh^2) + (1/3*M(rod)*L(rod)^2)} / (F*g*r*sin @)

Explanation / Answer

torque is L = (12.1m). 18kg.9.8m/s^2 sin 35^0 = I . angular acceleration.

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