A 3kg rod that it 2.3 m long is free to rotate in a vertical plane about an axle

Question

A 3kg rod that it 2.3 m long is free to rotate in a vertical plane about an axle, that runs through e rods center is perpendicular to the rod's length, and runs parallel to the floor. A 1kg block is attached to one end of the rod and a 2kg block is attached to the other end at some instant, the rod makes an angle of 30 degree with the horizontal so that the blocks are in the positions shown in the figure. Ignore friction and assume the blocks are small enough that any length they add to the rod can be ignored. Determine the torque caused by the forces exerted on the system at this instant Determine the rotational acceleration of the system at this instant. which weighs 200N rests on a wall (as shown in the figure). The ladder the bottom. There is friction on the ground where the coefficient of friction along the wall where the coefficient of friction is 0.04 How far along

Explanation / Answer

3.

As the axle is through the middle the rod applies no torque. Only the weights on the end.
The distance of each weight, from the middle, perpendicular to gravity, is 2.3 / 2 * cos(30)
So the torque must be mx
= (2 - 1 ) * 2.3 / 2 * cos(30) = 0.996 N m

b) the moment of inertia is made of three parts.
1) the moment of inertia of the rod. You should know the formula. I = mL^2/ 12= 3 * 1/3 * (2.3/2)^2
= 1* (2.3/2)^2
2) the moment of inertia of the 1 kg mass = 1 * (2.3/2) ^2
3 the moment of inertia of the 2 kg mass = 2 * (2.3 / 2)^2

Add them up to get I = 4 * (2.3 / 2)^2 = 2.3^2

Now acceleration = torque / moment = 0.996 / ( 2.3^2) = 0.188 rad/ s^2

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