If instead ofthe force F an actual mass m = 0.750 kg is hung from the string, fi

Question

If instead ofthe force F an actual mass m = 0.750 kg is hung from the string, find the angular acceleration of the cylinder. 74.4 radjs 2 Hint: The tension in the string induces the torque in both this part and the first part. The tension is not equal to mq! If it were, the mass would not accelerate downward. Determine all of the forces acting on the mass, then apply Newton's second law and solve for the tension, and apply it to Newton's second law of rotational motion. Submit Answer You have entered that answer before Incorrect. Tries 3/12 Previous Tries How far does m travel downward between 0.650 s and 0.850 s after the motion begins? Submit Answer Tries 0/1:2 The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.445 m in a time of 0.510 s. Find Icm of the new cylinder

Explanation / Answer

Solution:

The tension is the force that produces torque to move the cylinder

so;

I = Torque

=> tension * radius = (MR2/ 2)

=> 7.357* 0.113 = {1.75*(0.113)2/2}

=> = 74.4 rad/s2

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