Question

A cord is wrapped around a pulley that is shaped like a disk of mass m and radius r. The cord’s free end is connected to a block of mass M. The block starts from rest and then slides down an incline that makes an anglewith the horizontal as shown in Figure P10.92. The coefficient of kinetic friction between block and incline is m.

(a) Use energy methods to show that the block’s speed as a function of position d down the incline is

$\mathit{v}\mathbf{=}\sqrt{\frac{\mathbf{4}\mathbf{M}\mathbf{g}\mathbf{d}\mathbf{(}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{-}\mathbf{\mu }\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{)}}{\mathbf{m}\mathbf{+}\mathbf{2}\mathbf{M}}}$

(b) Find the magnitude of the acceleration of the block in terms of,$\mathit{\mu }$, m,M, g and $\mathit{\theta }$.

Short answer

The solution is

1. $v_d=\sqrt{\left[\frac{4gdM(\sin \theta -\mu \cos \theta )}{m+2M}\right]}$
   
2. $a=2g\left(\frac{{\displaystyle M}}{{\displaystyle m+2M}}\right)(\sin \theta -\mu \cos \theta )$
   

Step-by-step solution

Given information

It is given that

Mass of the pulley =m

Radius of the pulley =r

Mass of the block =M

Angle of inclination = $\theta$

Coefficient of kinetic friction between the block and the incline =M

Now, the final kinetic energy$K_f=\frac{1}{2}M{v}_f^2+\frac{1}{2}I{\omega }_f^2$

Final potential energy$U_f=mg{h}_f=0$

Initial kinetic energy$K_i=0$

Initial potential energy$U_i={\left(Mgh\right)}_i$

$\begin{array}{c}f=\mu N\\ =\mu Mg\cos \theta \\ \omega =\frac{v}{r}\\ h=d\sin \theta \end{array}$

And,

$I=\frac{1}{2}m{r}^2$

The block’s speed using the energy methods

(a)

The energy shift will be.

$\begin{array}{l}\Delta E=E_f-E_i\\ -fd=K_f+U_f-K_i-U_i\\ -fd=\frac{1}{2}M{v}_f^2+\frac{1}{2}I{\omega }_f^2-Mgh\end{array}$

By substituting the values we get,

$\begin{array}{c}-\left(\mu Mg\cos \theta \right)d=\frac{1}{2}M{v}^2+\left(\frac{m{r}^2}{2}\right)\left(\frac{v^2}{r^2}\right)-Mgd\sin \theta \\ \frac{1}{2}[M+\frac{m}{2}]v^2=Mgd\sin \theta -\left(\mu Mg\cos \theta \right)d\\ v^2=2Mgd\frac{(\sin \theta -\mu \cos \theta )}{(\frac{m}{2}+M)}\end{array}$

As a result, the block speed as a function of incline position will be

$v_d=\sqrt{\left(\frac{4gdM(\sin \theta -\mu \cos \theta )}{m+2M}\right)}$

The magnitude of the acceleration of the block

(b)

We utilize the equation of motion, which is given by to find the acceleration.

$\begin{array}{c}{v}_f^2=v_i^2+2a\Delta x\\ v_d^2=2ad\\ a=\frac{v_d^2}{d}\\ =2g\left(\frac{M}{m+2M}\right)(\sin \theta -\mu \cos \theta )\end{array}$