Question

Disk A has a mass of $4$kg and a radius $r=75mm,$ it is at rest. when it is placed in contact with the belt. which moves at a constant speed $v=18m/s$. Knowing that ${\mu }_k=0.25$ between the disk and the belt, determine the number of revolutions executed by the disk before it reaches a constant angular velocity. (Assume that the normal reaction by the belt on the disc is equal to weight of the disc).

Answer 1

R.E.F image

Mars of disc $M=4Kg$

Radius of disc $R=75\times {10}^{-3}m$

Speed of belt , $V=18m/s$

Coeff of kinetic fractious $f_r=0.25$

When the belt comes in contact with the disc

there is relative motion between the surface

This causes kinetic friction to act

The nominal force between belt and disc is equal is equal to the

weight.

So, FBD of disc is

Torque due to friction $\tau =f_r\times R$

$=0.25\times 40\times 75\times {10}^{-3}$

$=0.75Nm$

Applying newtons laws $\tau =T\alpha$

$=0.75=\frac{M{R}^2}{2}\times \alpha$

$=\frac{0.75}{100}=\frac{4\times 5625\times {10}^{-6}}{2}\alpha$

$\alpha =\frac{2\times {10}^4}{4\times 75\times 100}=\frac{2\times 100}{300}=66.66\%$

$W=\frac{V}{r}=\frac{18}{75}\times {10}^{-3}=\frac{8000}{75}=240$

$w^2=2\alpha \theta$

$\theta =\frac{{240}^2}{66.66}=137.52$