Question

A solid ball of inertia m rolls without slipping down a ramp that makes an angle θ with the horizontal.

Answer 1

Answer:

Part a)

$f = \frac{2}{7}mgsin\theta$

Part b)

$\mu = \frac{2}{7} gtan\theta$

Explanation:

Part a)

Force equation on the inclined plane is given as

$mgsin\theta - f = ma$

now for torque equation of the ball

$fR = \frac{2}{5}mR^2 ( \frac{a}{R})$

$f = \frac{2}{5}ma$

now from above two equations

$mg sin\theta - \frac{2}{5}ma = ma$

$mg sin\theta = \frac{7}{5} ma$

$a = \frac{5}{7} gsin\theta$

so frictional force is given as

$f = \frac{2}{5}ma$

$f = \frac{2}{5}m(\frac{5}{7}gsin\theta)$

$f = \frac{2}{7}mgsin\theta$

Part b)

Also we know that in the normal direction of the motion we have

$F_n = mgcos\theta$

so we have

$f = \mu F_n$

$\frac{2}{7} mg sin\theta = \mu (mg cos\theta)$

now we have

$\mu = \frac{2}{7} gtan\theta$