(C) The impulse $J$ applied at one end provides both linear and angular momentum to the system.$1$. Linear impulse equation: $J = M_{total} v_{cm} \Ri

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(C) The impulse $J$ applied at one end provides both linear and angular momentum to the system.$1$. Linear impulse equation: $J = M_{total} v_{cm} \Rightarrow MV = (2M) v_{cm} \Rightarrow v_{cm} = \frac{V}{2}$.$2$. Angular impulse equation about the center of mass $(COM)$: $J \times r = I_{cm} \omega$.Here,$r = \frac{L}{2}$ and $I_{cm} = M(\frac{L}{2})^2 + M(\frac{L}{2})^2 = \frac{ML^2}{2}$.Substituting the values: $(MV) \times \frac{L}{2} = (\frac{ML^2}{2}) \omega$.Solving for $\omega$: $\frac{MVL}{2} = \frac{ML^2 \omega}{2} \Rightarrow \omega = \frac{V}{L}$.

Class 11 Physics System of Particles and Rotational Motion Relation between Torque and Angular acceleration and it's Application

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Consider a body shown in the figure,consisting of two identical balls,each of mass $M$,connected by a light rigid rod of length $L$. If an impulse $J = MV$ (where $V$ is the linear velocity of a ball) is imparted to the body at one of its ends,what would be its angular velocity?

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A
$\frac{V}{3L}$
B
$\frac{V}{4L}$
C
$\frac{V}{L}$
D
$\frac{2V}{L}$

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Relation between Torque and Angular acceleration and it's Application

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Consider a body shown in the figure,consisting of two identical balls,each of mass $M$,connected by a light rigid rod of length $L$. If an impulse $J = MV$ (where $V$ is the linear velocity of a ball) is imparted to the body at one of its ends,what would be its angular velocity?

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$A$ solid sphere of mass $2 \,kg$ and radius $1 \,m$ is free to rotate about an axis passing through its centre. $A$ constant tangential force $F$ is required to rotate the sphere with $10 \,rad/s$ in $2 \,s$ starting from rest. The value of $F$ is . . . . . . (in $\,N$)

An electric motor is started from rest to produce a torque. It acquires an angular acceleration given by $\alpha = 3t - t^2$. The angular acceleration becomes zero after $2 \ s$. Calculate the angular velocity after $6 \ s$.

$A$ constant torque of $400 \ Nm$ acts on a wheel having a moment of inertia of $100 \ kg \ m^2$ about its central axis. The angular velocity acquired in $4 \ s$ is ....... $rad \ s^{-1}$.

$A$ wheel is at rest in a horizontal position. Its moment of inertia about the vertical axis passing through its centre is $I$. $A$ constant torque $\tau$ acts on it for $t$ seconds. The change in rotational kinetic energy is:

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