Finding out torque in a viscosimeter

Question

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

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.

Answer 1

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.

Finally:

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

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

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

Answer 2

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.