A wheel which is initially at rest is subject | vedclass

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

Question

(B) Given that the wheel starts from rest,the initial angular velocity $\omega_0 = 0$. The angular displacement $	heta$ under constant angular accele

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) Given that the wheel starts from rest,the initial angular velocity $\omega_0 = 0$. The angular displacement $	heta$ under constant angular acceleration $\alpha$ is given by $	heta = \omega_0 t + \frac{1}{2} \alpha t^2$.For the first time interval $t$,the angle rotated is $	heta_1 = 15^{\circ}$.$15^{\circ} = 0 + \frac{1}{2} \alpha t^2 \implies \frac{1}{2} \alpha t^2 = 15^{\circ} \quad \dots(i)$For the total time interval $(t + 2t) = 3t$,the total angular displacement $	heta_{total}$ is:$	heta_{total} = \frac{1}{2} \alpha (3t)^2 = \frac{1}{2} \alpha (9t^2) = 9 \left( \frac{1}{2} \alpha t^2 ight)$.Substituting the value from equation $(i)$:$	heta_{total} = 9 	imes 15^{\circ} = 135^{\circ}$.The increase in angle in the next $2t \ s$ is $\Delta 	heta = 	heta_{total} - 	heta_1 = 135^{\circ} - 15^{\circ} = 120^{\circ}$.

$A$ wheel which is initially at rest is subjected to a constant angular acceleration about its axis. It rotates through an angle of $15^{\circ}$ in time $t \ s$. The increase in angle through which it rotates in the next $2t \ s$ is (in $^{\circ}$)

Similar Questions

If $\frac{1-x+6x^2}{x-x^3} = \frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x}$,then $A$ is equal to

The solution of $\frac{dy}{dx} = \frac{x \log x^2 + x}{\sin y + y \cos y}$ is

Which of the following is included in biochemistry?

The catalyst used for the polymerization of olefins is....

Vedclass Test Series

Exam Paper Generator

Online Exam Module