The instantaneous angular position of a point | vedclass

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

Question

(A) Given the angular position: $	heta(t) = 2t^3 - 6t^2$. First,find the angular velocity $\omega$ by differentiating $	heta$ with respect to time $

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the angular position: $	heta(t) = 2t^3 - 6t^2$. First,find the angular velocity $\omega$ by differentiating $	heta$ with respect to time $t$: $\omega = \frac{d	heta}{dt} = 6t^2 - 12t$. Next,find the angular acceleration $\alpha$ by differentiating $\omega$ with respect to time $t$: $\alpha = \frac{d\omega}{dt} = 12t - 12$. The torque $	au$ is related to angular acceleration by the equation $	au = I\alpha$,where $I$ is the moment of inertia. For the torque to be zero,the angular acceleration $\alpha$ must be zero (assuming $I 
eq 0$). Setting $\alpha = 0$: $12t - 12 = 0$. Solving for $t$: $12t = 12$,which gives $t = 1 	ext{ s}$.

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t) = 2t^3 - 6t^2$. The torque on the wheel becomes zero at $t = $ ....... $\text{sec}$.

Similar Questions

Resonance is an example of

The complete set of values of $x$ for which the function $f(x) = 2 \log_e(x-2) - x^2 + 4x + 1$ is increasing is:

$S$ and $T$ are the foci of an ellipse and $B$ is an end point of the minor axis. If $\triangle STB$ is an equilateral triangle, then the eccentricity of the ellipse is:

In $YDSE$,the separation between the slits is halved and the distance between the slit and the screen is doubled. The fringe width is

When a vibrating tuning fork is placed on a sound box of a sonometer, $8$ beats per second are heard when the length of the sonometer wire is kept at $101 \, cm$ or $100 \, cm$. Then the frequency of the tuning fork is (Consider that the tension in the wire is kept constant). (in $ \, Hz$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module