(C) In linear motion,work done is given by $W = F \Delta x$. The rotational analogue of force $F$ is torque $\tau$,and the analogue of displacement $\

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(C) In linear motion,work done is given by $W = F \Delta x$. The rotational analogue of force $F$ is torque $\tau$,and the analogue of displacement $\Delta x$ is angular displacement $\Delta \theta$. Thus,the rotational work is $W = \tau \Delta \theta$. This corresponds to $(1-b)$.In linear motion,power is given by $P = Fv$. The rotational analogue of force $F$ is torque $\tau$,and the analogue of velocity $v$ is angular velocity $\omega$. Thus,the rotational power is $P = \tau \omega$. This corresponds to $(2-a)$.Therefore,the correct matching is $(1-b), (2-a)$.

Class 11 Physics System of Particles and Rotational Motion Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

Easy

Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$

$(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$

$(c)$ $L = I \omega$

A
$(1-c), (2-a)$
B
$(1-b), (2-c)$
C
$(1-b), (2-a)$
D
$(1-a), (2-b)$

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System of Particles and Rotational Motion

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Energy conservation, Kinetic Energy, Work and Power of Rotational Motion

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$A$ tangential force $F$ is applied on the rim of a ring of radius $R$. If the ring rotates by an angle $\theta$,what is the work done by the force?

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Column-$I$	Column-$II$
$(1)$ $W = F \Delta x$	$(a)$ $P = \tau \omega$
$(2)$ $P = Fv$	$(b)$ $W = \tau \Delta \theta$
	$(c)$ $L = I \omega$

Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$. Column-$I$ Column-$II$ $(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$ $(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$ $(c)$ $L = I \omega$

Similar Questions

$A$ thin,uniform metal rod of mass $M$ and length $L$ is swinging about a horizontal axis passing through its end. Its maximum angular velocity is $\omega$. Its centre of mass rises to a maximum height of $(g = \text{acceleration due to gravity})$

$A$ thin rod of length $l$ and mass $m$ oscillates in a vertical plane about a horizontal axis passing through one of its ends. If the maximum angular velocity of the rod is $\omega$,what is the maximum height reached by its center of mass?

$A$ uniform rod of length $2L$ is placed with one end in contact with a horizontal surface. The other end is released from an angle $\alpha$ with the horizontal,such that the end in contact does not slip. What will be its angular velocity when it becomes horizontal?

$A$ thin uniform rod of length $L$ and mass $M$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of (where $g$ is the acceleration due to gravity):

$A$ tangential force $F$ is applied on the rim of a ring of radius $R$. If the ring rotates by an angle $\theta$,what is the work done by the force?

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Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$.

Column-$I$ Column-$II$

$(1)$ $W = F \Delta x$ $(a)$ $P = \tau \omega$

$(2)$ $P = Fv$ $(b)$ $W = \tau \Delta \theta$

$(c)$ $L = I \omega$