The M.I. of a body about the given axis is 1. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given: Moment of Inertia $I = 1.2 \, kg \cdot m^2$,Initial angular velocity $\omega_0 = 0$,Rotational Kinetic Energy $K_r = 1500 \, J$,Angular acc

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given: Moment of Inertia $I = 1.2 \, kg \cdot m^2$,Initial angular velocity $\omega_0 = 0$,Rotational Kinetic Energy $K_r = 1500 \, J$,Angular acceleration $\alpha = 25 \, rad/s^2$.Step $1$: Calculate the final angular velocity $\omega$ using the formula for rotational kinetic energy:$K_r = \frac{1}{2} I \omega^2$$1500 = \frac{1}{2} 	imes 1.2 	imes \omega^2$$1500 = 0.6 	imes \omega^2$$\omega^2 = \frac{1500}{0.6} = 2500$$\omega = \sqrt{2500} = 50 \, rad/s$.Step $2$: Calculate the time $t$ required using the kinematic equation $\omega = \omega_0 + \alpha t$:$50 = 0 + 25 	imes t$$t = \frac{50}{25} = 2 \, s$.

The $M.I.$ of a body about the 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 duration of ........ $s$.

Similar Questions

The power $(P)$ is supplied to a rotating body having moment of inertia '$I$' and angular acceleration '$\alpha$'. Its instantaneous angular velocity is

Explain the work done by torque.

$A$ rod of length $1 \ m$ is held vertically and allowed to fall such that its lower end remains fixed on the ground. The velocity of the other end of the rod when it hits the ground is ......... $m/s$ $(g = 9.8 \ m/s^2)$.

$A$ uniform rod of mass $M$ and length $L$ is pivoted at one end and is free to rotate in a vertical plane. The rod is released from rest in a horizontal position. What is the angular velocity of the rod when it reaches the vertical position?

Vedclass Test Series

Exam Paper Generator

Online Exam Module