Question

A sphere rolls up an inclined plane of inclination angle 30 degree. At the bottom of the incline the center of mass of the sphere has a translational speed 0.25 m/s. How far does the sphere travel up the plane

Answer 1

Answer: The sphere is 0.002548 cm travel up the plane.

Step-by-step explanation:

Since we have given that

Inclination angle = 30°

Translational speed = 0.25 m/s

As we know that

$KE=0.5\times lw^2=PE$

and

Length of solid sphere is given by

$l=\dfrac{2Mr^2}{5}$

So, it becomes,

$KE=0.5\times \dfrac{2Mr^2}{5}w^2=Mgd\sin \theta$

And $0.25\ m/sv=rw$

So, it becomes,

$1\times (0.25)^2=5\times d\times 9.81\times \sin 30^\circ\\\\0.0625=49.05d\times \dfrac{1}{2}\\\\\dfrac{0.125}{49.05}=d\\\\d=0.002548$

Hence, the sphere is 0.002548 cm travel up the plane.