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A thin uniform rod of length 3 m and mass M is held horizontally by two vertical strings attached to the two ends, as shown in figure. One of the strings is cut. Find the angular acceleration of the rod at the moment it is cut. Take g=10 m/s2.
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Q: A one uniform rod of mass m and length L is held horizontally by two vertical strings of negligible mass as shown in the figure. Immediately after the string is cut, then
(a) liner acceleration of free end of the rod is (3/2)g
(b) liner acceleration of centre of mass of the rod is (3/4)g
(c) tension in the left string is mg/2
(d) tension in the left string is mg/2
Ans: (a), (b), (c)
Answer
Hint: When the string is cut, the rod will rotate around the other end which is tied to the string. The torque responsible for the rotation is due to the weight of the rod at the centre of the rod. Use the relation between torque and moment of inertia to solve this question.
Formula used:
The torque is given as,\[\tau = r\,F\sin \theta \]Here, F is the force that causes rotation, r is the position of point of force from the centre of rotation and \[\theta \] is the angle between length of the rod and direction of force.The relation between torque and moment of inertia is,\[\tau = I\alpha \]Here, I is the moment of inertia of the rod and \[\alpha \] is the angular acceleration.Complete step by step answer:
After one of two strings is cut, the rod will move in a circular path around its other end tied to the string as shown in the figure below.
So, the correct answer is “Option C”.
Note:
The tension in the string also acts on string as one of the forces, but the torque is zero when the force acts at the centre. The angular acceleration and linear acceleration of the rod is different. The linear acceleration is the tangential acceleration of the rod given by \[\alpha L\].