You have a force of 4mg pulling on the apparatus. You know the torque, given r.
Torque= I*angular acceleration.
You know I (you have each m, and L) Solve for angularacceleration.
Now that you have angular acceleration, solve for tangential acceleration,given r. Isn't that the same acceleration of the downward block? Wont the total KE be equal to the change in PE?
A light string that is attached to a large block of mass 4m passes over a pulley with negligible rotational inertia and is wrapped around a vertical pole of radius r. The system is released from rest, and as the block descends the string unwinds and the vertical pole with its attached apparatus rotates. The apparatus consists of a horizontal rod of length 2L, with a small block of mass m attached at each end. The rotational inertia of the pole and the rod are negligible. A) Determine the rotational inertia of the rod-and-block apparatus attached to the top of the pole. B) Determine the downward acceleration of the large block. C) When the large block has descended a distance D, how does the instantaneous total kinetic energy of the three blocks compare with the value 4mgD? (greater, equal, less)----I don't even know where to begin here, I am so confused...any help with any part of this, even just getting me started would be very much appreciated!
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The force pulling downwards is 4mg but you must take into account Tension. so Torque = I*angular acceleration which in turn ='s rF. where F = Tension(T).
1 answer