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Work, Energy And Power For Rotating Body MCQ - Practice Questions with Answers
Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET
Quick Facts
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Work, Energy and Power for Rotating Body is considered one of the most asked concept.
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14 Questions around this concept.
Solve by difficulty
Two rotating bodies A and B of masses m and 2m with moments of inertia IA and have equal kinetic energy of rotation. If LA and LB are their angular momenta respectively, then
A body of radius $R$ and mass $M$ is initially rolling on a level surface with speed $\omega$. It then rolls up an inclined to a maximum height $h$. If $h=\frac{\mathrm{u}^2}{\mathrm{~g}}$. The geometrical shape of the body is.
A hollow sphere rolling on a surface what is the % of its translation kinetic energy-
A solid sphere is rolling on a surface its speed is $(\mathrm{V})$ and mass $(\mathrm{M})$. The its kinetic energy is
when a sphere of the moment of inertia.$I$ rolls down on an inclined plane then the percentage of rotational kinetic energy will be -
A dog of mass 10kg runs at a speed of 5m/sec at the surface of a circular drum of radius 1m. The drum can rotate about its axis AB freely as shown in the figure. The motion of the dog causes the rotation of the drum. The speed of rotation is such that the relative position of the dog is unaltered. The kinetic energy of the rotation of a dog is
An object of mass '2kg' is rotating with an angular velocity of '1rad/s' about an axis that is '1m' away from the centre of mass. If the radius of gyration of this object is '2m'. Then rotational kinetic energy of the body is given by:
Concepts Covered - 1
Work, Energy and Power for Rotating Body
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Work-
For translation motion
$$
W=\int F d s
$$
So for rotational motion
$$
W=\int \tau d \theta
$$
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Rotational kinetic energy-
The energy a body has by virtue of its rotational motion is called its rotational kinetic energy.
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Rotational kinetic energy |
Translatory kinetic energy |
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1 |
$K_R=\frac{1}{2} I \omega^2$
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$K_T=\frac{1}{2} m V^2$
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2 |
$K_R=\frac{1}{2} I \omega$
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$K_T=\frac{1}{2} P V$
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3 |
$K_R=\frac{L^2}{2 I}$
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$K_T=\frac{P^2}{2 m}$
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Power =Rate of change of kinetic energy
For translation motion $P=\vec{F} \cdot \vec{V} P=\vec{F} \cdot \vec{V}$
So for rotational motion
$$
P=\frac{d\left(K_R\right)}{d t}=\frac{d\left(\frac{1}{2} I \omega^2\right)}{d t}=I \omega \frac{d \omega}{d t}=I \alpha \omega=\tau \cdot \omega
$$
Or $P=\vec{\tau} \cdot \vec{\omega}$
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