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A bob of mass $m$ is attached to the base of a cart via a thread of length $l$. The cart is accelerating with a non-uniform acceleration with magnitude $a$. The positive x-direction is towards the right and the positive y-direction is downwards. A simple diagram of the situation is shown in the figure above. The problem which I am facing is that when I calculate the magnitude of tension $T$ in the thread using two different approaches, I get two different values. Please help me find the mistake in my solution.

First approach (Observer is stationary)

Assuming the magnitude of the displacement of the cart from its initial position to be $x_c$, the bob's $x$ and $y$ displacement from its initial position can be written as:
$x = x_c + l\sin\theta$
$y = l\cos\theta$

From Newton's laws, we will get:
$-T\sin\theta = m\ddot x = m\ddot x_c + ml(-\dot \theta^2\sin\theta + \ddot \theta\cos\theta)$
$mg - T\cos\theta = m\ddot y = ml(-\dot \theta^2\cos\theta - \ddot \theta\sin\theta)$

If we multiply the above equations by $\sin\theta$ and $\cos\theta$ respectively and add them up, we get the following equation: $T = mg\cos\theta -ma\sin\theta + ml\dot \theta^2$

Second approach (Observer is present in the cart)
From the point of view of this observer, a pseudo force will act on the bob in the negative x-direction. From this reference frame, since there will be no displacement of the bob along the thread, from Newton's laws: $T = mg\cos\theta - ma\sin\theta$

As you can see, we obtain two different values of $T$ using two different approaches. Please help me find out the mistake I am making.

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Your first approach is correct.

For the second approach, You have not taken into account centripetal acceleration $m\omega^2l$ which is the term you are missing in your second equation.

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  • $\begingroup$ @ahamshubham When you whirl a stone that's attached to a rope with your hand in the horizontal plane, the stone doesn't displace along the direction of the taut rope and yet we know that there's a net non-zero force (the tension of the rope) that's responsible for the centripetal acceleration experienced by the stone. $\endgroup$
    – Ajay Mohan
    Nov 4 '19 at 7:01
  • $\begingroup$ @user35508 Thank you for your answer. It clarifies everything. $\endgroup$
    – user198167
    Nov 4 '19 at 7:42
  • $\begingroup$ @AjayMohan Thanks for your explanation but I think there is a displacement in the direction of the thread which leads to the centripetal acceleration and this acceleration is the quantity which I did not consider in my equation. $\endgroup$
    – user198167
    Nov 4 '19 at 7:44
  • $\begingroup$ @ahamshubham Why do you say that? In the frame of the observer located in the cart, the mass $m$ doesn't experience a displacement along the direction of the thread since its velocity is perpendicular to the thread, right? The centripetal acceleration is just resposible for changing the direction of the velocity of $m$. $\endgroup$
    – Ajay Mohan
    Nov 4 '19 at 8:06
  • $\begingroup$ @AjayMohan Oh, yeah. Makes perfect sense. $\endgroup$
    – user198167
    Nov 4 '19 at 22:59

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