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If a rod of mass m is fixed at one end and a particle of mass m is attached to its other end,...

Question:

If a rod of mass m is fixed at one end and a particle of mass m is attached to its other end, then what is the minimum velocity required for the system to perform circular motion?

Why is the moment of inertia of the rod taken without any influence of particle and, the Rotational energy and Kinetic Energy (KE) is initial energy in the energy conservation equation. Explain Why?

Rotational Kinetic Energy:

For the minimum velocity at the bottom, the final angular velocity at the top must be zero. The Rotational kinetic energy with the minimum angular velocity must be equal to the change in potential energy of the rod at the highest point. The Moment of inertia of the system about the axis of rotation is the sum of moment of inertia of rod about its end and the moment of inertia of particle,

Answer and Explanation:

Position of center of mass from the point of suspension

{eq}OC= \dfrac{m(0.5L)+m(L)}{m+m} \\ OC= \dfrac{3}{4}L {/eq}

Now the system can be considered as a point mass rotating in a vertical circle of radius {eq}R=0.75L {/eq}

Rise in center of mass from bottom to top position

{eq}h=2 \times 0.75L \\ h=1.5L {/eq}

Moment of inertia of the system about the point of suspension

{eq}I= \dfrac{1}{3}mL^2 +mL^2 \\ I=\dfrac{4}{3}mL^2 {/eq}

For the minimum angular velocity

Rotational KE at the lowest point = PE at the top most point

{eq}\dfrac{1}{2}I \omega^2 =mgh \\ \dfrac{1}{2} \times \dfrac{4}{3}mL^2 \times \omega^2 =mg (1.5L) \\ \omega = \dfrac{1.5g}{L} {/eq}


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Kinetic Energy of Rotation

from UExcel Physics: Study Guide & Test Prep

Chapter 7 / Lesson 7
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