(B) Given: Moment of Inertia $I = 1.2 \ kg \cdot m^2$,Rotational Kinetic Energy $K_r = 1500 \ J$,Angular acceleration $\alpha = 25 \ rad/s^2$,Initial

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(B) Given: Moment of Inertia $I = 1.2 \ kg \cdot m^2$,Rotational Kinetic Energy $K_r = 1500 \ J$,Angular acceleration $\alpha = 25 \ rad/s^2$,Initial angular velocity $\omega_0 = 0$.The formula for rotational kinetic energy is $K_r = \frac{1}{2} I \omega^2$.Substituting the values: $1500 = \frac{1}{2} \times 1.2 \times \omega^2$.$1500 = 0.6 \times \omega^2 \Rightarrow \omega^2 = \frac{1500}{0.6} = 2500$.Therefore,the final angular velocity $\omega = \sqrt{2500} = 50 \ rad/s$.Using the kinematic equation for rotation: $\omega = \omega_0 + \alpha t$.$50 = 0 + 25 \times t$.$t = \frac{50}{25} = 2 \ s$.

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

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The $M.I.$ of a body about a given axis is $1.2 \ kg \cdot m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \ J$,an angular acceleration of $25 \ rad/s^2$ must be applied about that axis for a time duration of $....... \ s$.

A
$4$
B
$2$
C
$8$
D
$10$

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Class 11

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System of Particles and Rotational Motion

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

$A$ 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?

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$A$ constant power is supplied to a rotating disc. The angular velocity $(\omega)$ of the disc varies with the number of rotations $(n)$ made by the disc as:

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$A$ disc with a moment of inertia of $\frac{9.8}{\pi^2} \ kg \ m^2$ is rotating at $600 \ rpm$. If the rotational frequency changes from $600 \ rpm$ to $300 \ rpm$,the work done is .......... $J$.

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The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?

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$A$ uniform rod of length $L$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $\theta$,its angular velocity $\omega$ is given as

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The $M.I.$ of a body about a given axis is $1.2 \ kg \cdot m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \ J$,an angular acceleration of $25 \ rad/s^2$ must be applied about that axis for a time duration of $....... \ s$.

Similar Questions

$A$ 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?

$A$ constant power is supplied to a rotating disc. The angular velocity $(\omega)$ of the disc varies with the number of rotations $(n)$ made by the disc as:

$A$ disc with a moment of inertia of $\frac{9.8}{\pi^2} \ kg \ m^2$ is rotating at $600 \ rpm$. If the rotational frequency changes from $600 \ rpm$ to $300 \ rpm$,the work done is .......... $J$.

The moment of inertia of a body about a given axis is $2.4 \ kg \cdot m^2$. To produce a rotational kinetic energy of $750 \ J$,an angular acceleration of $5 \ rad/s^2$ must be applied about that axis for how many seconds?

$A$ uniform rod of length $L$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $\theta$,its angular velocity $\omega$ is given as

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