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A thin ring of mass M and radius R is rolling down an inclined plane of inclination angle \theta. As ring moves down the incline, it starts winding a thin ribbon of linear mass density \lambda. If the height of the incline is H. then the distance it will travel from the foot of the incline before coming to rest will be.

\begin{aligned} & {\left[\text { Take, } M=0.1 \mathrm{~kg}, R=1 \mathrm{~m}, H=1 \mathrm{~m}, \theta=30^{\circ},\right.} \\ & \left.\lambda=1 \mathrm{~kg} / \mathrm{m}, g=10 \mathrm{~m} / \mathrm{s}^2\right] \end{aligned}

Option: 1

2.5 m


Option: 2

1.5 m


Option: 3

5.0 m


Option: 4

2.0 m


Answers (1)

best_answer

Initial energy of the system just before the ring. starts on the inclined plane.
E_i=M g(R+H)+\frac{\lambda g H^2}{2 \sin \theta}
As friction is not present, so when the ring stops; the total energy well be potential energy.
E_f=M g R+\lambda g R\left(\frac{H}{\sin \theta}\right)+\lambda g R(S)

Using energy conservation,         \begin{aligned} & E_i=E_f \\ & M g(R+H)+\frac{\lambda g H^2}{2 \sin \theta}=M g R+\lambda g R\left[\frac{H}{\sin \theta}+S\right] \\ & S=\frac{M g+\lambda\left(\frac{H}{\sin \theta}\right)\left(R-\frac{H}{2}\right)}{\lambda R} \\ & S=\frac{(0.1 \times 10)+(1) \frac{(1)}{(1 / 2)}\left(1-\frac{1}{2}\right)}{1 \times 1} \\ & S=\frac{1+1}{1}=2 \mathrm{~m} \end{aligned}    

Posted by

sudhir.kumar

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