This is a problem from my introductory physics textbook:

A wheel of moment of inertia $I$ and radius $r$ is free to rotate about its centre. A string is wrapped over its rim and a block of mass m is attached to the free end of the string. The system is released from rest. Find the speed of the block as it descends through a height $h$.

The answer in my book utilizes energy considerations, reasoning that "the gravitational potential energy lost by the block must be equal to the kinetic energy gained by the block and the wheel." Thus,

$$mgh=\frac{mv^2}{2}+\frac{Iv^2}{2r^2}$$ $$\rightarrow v=\sqrt{\frac{2mgh}{m+I/r^2}}$$

My question is, is it possible to solve this question using only the equations of kinematics of rotational motion, viz. $$\omega = \omega_0+\alpha t$$ $$\Delta \theta=\omega_0t+1/2\alpha t^2$$ $$\omega^2=\omega_0^2+2\alpha \Delta \theta$$

As an analogy, consider the case when the wheel was massless. Then, the equation from energy considerations would have been: $$mgh=\frac{mv^2}{2}$$

Solving the above, we get, $$v=\sqrt{2gh}$$

When we use the equation $v^2=u^2+2gh$, setting $u=0$ for the system starting from rest, we again get, $v=\sqrt{2gh}$.


2 Answers 2


The case with a massless wheel is a bit boring, because it is just the case without a wheel, i.e., a free falling body.

In classical mechanics, there is always multiple routes that lead to the same answer, so you could start from the kinematics equations if you can argue that the angular acceleration $\alpha$ is constant. However, if you don't start from the energy point of view, you have to work from a force point of view. So, what you have to do, is draw the forces on your sketch (you have a sketch right?). Then, what you need is the tension in the string. The tension force $T$ exerted by the string on the mass, is the same as the tension force exerted by the string on the wheel (why?).

Since this is a homework question, I'll leave the rest to you (hint: determine $\alpha$ based on what you know about $T$). The force and energy approach will lead to the same answer. However, in some cases, one approach is much easier to analyze than the other.

  • $\begingroup$ I got $a=Tr/I$, where $a$ is the linear acceleration of a point on the rim of the wheel, and therefore, the block's acceleration. Could you give me a hint on how to find $T$? $\endgroup$
    – Polemos
    Nov 12, 2020 at 18:21
  • $\begingroup$ You need one more expression for $T$ and $a$. $\endgroup$
    – Bernhard
    Nov 12, 2020 at 18:24
  • 1
    $\begingroup$ Thanks. I got it now. $\endgroup$
    – Polemos
    Nov 12, 2020 at 18:32

To write a more complete answer for other readers: I would use Newton's second law and its rotational equivalent. Use $F= ma$ for the block of mass $m$ (since there is only translation, and no rotation), and then use $\tau = I \alpha$ for the wheel (since there is only rotation about its center, and no translation).

The tension force exerted on the string as pointed out by Bernhard will be $T$, and this force acts on the mass (such that $\Sigma F_{\rm mass} = mg -T$) and also provides a torque on the wheel about its center, such that $\Sigma \tau = RT$. Since there is no slipping, use the fact that $a_{\rm mass} = R \alpha_{\rm wheel}$. System of equations, solve for $\alpha$.

Then, relate the distance travelled for the bloc, $h$, and the angular displacement by knowing that $h=R\theta$. After that you can use the $\theta = \tfrac{1}{2}\alpha t^2$ equation.


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