A solid sphere is resting over a rough horizontal floor. A sharp impulse is applied on it along its horizontal diameter. Point out false statements :

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The kinetic energy of the sphere remains constant throughout the motion

The kinetic energy of translation decreases and rotational kinetic energy increases, but the total kinetic energy decreases initially and finally attains a constant value

The total kinetic energy of the sphere decreases continuously on doing work against friction and finally reduces to zero

A constant frictional force, opposite to the translational motion acts on the sphere throughout its motion.

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Since, the impulse is applied along a horizontal diameter, therefore, due to that impulse the sphere starts to move translationally without any rotational motion. Since the floor is rough, therefore, friction comes into existence and that opposes forward sliding of the sphere as shown in figure. Hence, friction acts along backward direction which not only provides a retarding force but produces an accelerating moment also. Due to that moment the sphere experiences an angular acceleration. Hence, initially translational Rotational Dynamics "ROLLING" velocity of the sphere decreases but angular velocity increases Therefore, $K E$ of the sphere initially decreases. Rotational motion is accelerated till angular velocity ( $ω$ ) becomes equal to $\frac{v}{r}$ and then the friction disappears. Hence, then total energy of sphere remains constant. Therefore, option (2) alone is correct and rest options are incorrect.

Watch 3-min video & get full concept clarity