Question

A force F acts tangentially at top of a solid sphere of mass M kept on rough horizontal plane. If sphere rolls without slipping, acceleration of its center is: (10F/7M is answer. But dont know how. Please help.)

Answer 1

Answer on Question #46810, Physics, Mechanics | Kinematics | Dynamics

Question:

A force F acts tangentially at top of a solid sphere of mass M kept on rough horizontal plane. If sphere rolls without slipping, acceleration of its center is: (10F/7M is answer. But dont know how. Please help.)

Answer:

If sphere rolls without slipping:

$$\beta = \frac{a}{R}$$

where $\beta$ is angular acceleration, $a$ is acceleration of center of the sphere.

Newton’s laws of motion:

$$(F + F_{fr})R = I\beta$$ $$F - F_{fr} = ma$$

where $F_{fr}$ is force of friction, $I$ is moment of inertia (for solid sphere $I = \frac{2}{5}mR^2$)

Assuming $\beta = \frac{a}{R}$:

$$(F + F_{fr}) = \frac{I}{R^2}a$$ $$F - F_{fr} = ma$$

Therefore:

$$2F = \frac{I}{R^2}a + ma$$ $$a = \frac{2F}{\frac{I}{R^2} + m} = \frac{2F}{\frac{2}{5}m + m} = \frac{10F}{7m}$$

Answer: $\frac{10F}{7m}$

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