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Question: When a solid sphere rolls without slipping down an inclined plane making an angle \(\theta\) with ...

When a solid sphere rolls without slipping down an inclined plane making an angle θ\theta with the horizontal, the acceleration of its centre of mass is a. If the same sphere slides without friction, its acceleration a’ will be

A

72a\frac { 7 } { 2 } \mathrm { a }

B

57a\frac { 5 } { 7 } \mathrm { a }

C
D

52a\frac { 5 } { 2 } \mathrm { a }

Answer
Explanation

Solution

Acceleration of the solid sphere when it rolls without slipping down an inclined plane is.

a=gsin⁡ω1+IMR2a = \frac { g \sin \omega } { 1 + \frac { \mathrm { I } } { \mathrm { MR } ^ { 2 } } }

For a solid sphere, I=25MR2\mathrm { I } = \frac { 2 } { 5 } \mathrm { MR } ^ { 2 }

∴a=gsin⁡θ1+25=57 gsin⁡θ\therefore \mathrm { a } = \frac { \mathrm { g } \sin \theta } { 1 + \frac { 2 } { 5 } } = \frac { 5 } { 7 } \mathrm {~g} \sin \theta …(i)

Acceleration of the same sphere when it slides when it slides without friction down an same inclined plane is , …(ii)

Divide (ii) by (i), we get,

a′a=75\frac { \mathrm { a } ^ { \prime } } { \mathrm { a } } = \frac { 7 } { 5 } or a′=75aa ^ { \prime } = \frac { 7 } { 5 } \mathrm { a }