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  • A uniform spherical shell of mass  1 kg and  radius 1 m  is rolling with slipping on a level road. find the distance covered by the particle at the point of contact when the she

A uniform spherical shell of mass  1 kg and  radius 1 m  is rolling with slipping on a level road. find the distance covered by the particle at the point of contact when the shell completes half revolution.
 

Option: 1

R


Option: 2

2R


Option: 3

3R


Option: 4

4R


Answers (1)

best_answer

considering point  P on the rim of the shell, its resultant velocity will be given by 

U_R=\sqrt{\left(U_{c \cdot m}\right)^2+(r \omega)^2+2\left(U_{c \cdot m}\right)(r\omega ) \cos (180-\theta)} \\

U_R=\sqrt{\left(U_{c \cdot m}\right)^2+(r \omega)^2-2\left(U_{c \cdot m}\right)(r \omega) \cos \theta } \\U_R=\sqrt{\left(U_{c \cdot m}\right)^2+(r \omega)^2-2\left(U_{c \cdot m}\right)(r \omega) \cos \theta } \\

U_R=2 \,U_{c \cdot m} \sin \left(\frac{\theta}{2}\right) \\

Let the distance travelled be 'S' then 

S=\int_0^{\pi / \omega} 2 \,U_{c \cdot m} \sin \frac{\omega t}{2} d t \\
S=2 \,U_{c \cdot m}\left[-\frac{\cos \frac{\omega t}{2}}{\omega / 2}\right]_0^{\pi / \omega} \\

S=-\frac{U_{c \cdot m}}{\omega}\left[\cos \frac{\omega x}{2} \frac{\pi}{\omega}-\cos 0\right] \times 4 \\

S=-\frac{U_{c \cdot m}}{\omega} \times 4 \times-1 \\

S=\frac{4 U_{c \cdot m}}{\omega}=4 R

Posted by

HARSH KANKARIA

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