I am solving problems through Fox and McDonald.Here is the problem

enter image description here

I tried by finding out the torque due to viscous forces which is $$\tau_1=\mu(2\pi Rh)\frac{R\omega}{a}$$ and the torque due to mass m1 as $$\tau_2=m_1gR$$ Writing into equation $$\tau_2- \tau_1=m_2R^2 \frac{d\omega}{dt}$$ integrating this and using boundary condition $\omega=0$ at $t=0$ I got $$\omega=\frac{m_1ga}{2\pi Rh\mu}[1-exp(\frac{-2\pi \mu htR}{am_2})]$$ . However I am missing $m_1+m_2$ instead of $m_2$ in the exponential part.Any ideas? Thanks.


2 Answers 2


You didn't account for the acceleration of $m_1$.

Setting up a free body diagram on the weight shows:

$$m_1g - T = m_1a_y$$

Where $T$ is the tension in the rope, $a_y$ is the acceleration of the block.

This leads to the following corrections:

$$\tau_1=\mu(2\pi Rh)\frac{R\omega}{a}*R$$

(The original had a value for force, whereas we need a torque.) This is because the entire viscous force would operate at a distance R from the origin.

$$\tau_2=TR = (m_1g-m_1a_y)R=m_1gR-m_1R^2\frac{d\omega}{dt}$$

To account for the tension in the rope properly, the acceleration of $m_1$ must be considered.


$$\tau_2- \tau_1=m_2R^2 \frac{d\omega}{dt}$$ $$m_1gR-m_1R^2\frac{d\omega}{dt} - \mu(2\pi Rh)\frac{R\omega}{a}*R = m_2R^2 \frac{d\omega}{dt}$$ $$m_1g - \mu(2\pi Rh)\frac{R\omega}{a} = (m_1+m_2)R \frac{d\omega}{dt}$$

Which is easily solved into:

$$\omega = \frac{m_1ga}{2\mu\pi R^2h}(1-exp(\frac{-2\mu\pi Rht}{a(m_1+m_2)}))$$

Clearly when the exponential disappears, the maximum speed will settle as

$$\omega = \frac{m_1ga}{2\mu\pi R^2h}$$


Mark has already pointed out what you are missing, and this is the Lagrangian approach which also should give the same result - and it does.

The position of $m_1$ is $x_1 = R \ \theta(t)$

The kinetic energy of $m_1$: $KE_1=\frac{1}{2}m_1x_1'(t)^2 =\frac{1}{2} m_1 R^2 \theta '(t)^2$

The potential energy of $m_1$: $PE_1=-m_1 g x_1=-m_1 g R \ \theta (t)$

The kinetic energy of the $m_2$: $KE_2=\frac{1}{2}I\theta '(t)^2=\frac{1}{2} m_2 R^2 \theta '(t)^2$

The Lagrangian: $L=KE_1+KE_2-PE_1=\frac{1}{2} m_1 R^2 \theta '(t)^2+\frac{1}{2} m_2 R^2 \theta '(t)^2+m_1 g R \theta (t)$

The generalized force is the torque $\tau_2$: $Q=-\mu(2\pi Rh)\frac{R\omega}{a}R$

The equations of motion are $\frac{d }{d t}\frac{d L}{d \theta'(t)}-\frac{d L}{d \theta(t)}=Q$ which turn out as

$$ m_1 R^2 \theta ''(t)+m_2 R^2 \theta ''(t)- m_1 g R=-\frac{2 \pi h \mu R^3 \theta '(t)}{a} $$

Substituting $\theta '(t)=\omega (t)$ and simplifying we get

$$\left(m_1+m_2\right) R \omega '(t)+ \frac{2 \pi h \mu R^2 \omega (t)}{a}-m_1 g=0$$

and the solution with $\omega(0) = 0$ is $$\omega(t) =\frac{ m_1 g a}{2 \pi h \mu R^2} \left(1-e^{-\frac{2 \pi h \mu R t}{a \left(m_1+m_2\right)}}\right)$$

which agrees with the free-body diagram approach.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.