A pulley of radius 2, m is rotated about its | vedclass

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

Question

(D) The torque applied is $	au = rF = 2(20t - 5t^2) = 40t - 10t^2\, N\cdot m$.Using $	au = I\alpha$,we have $40t - 10t^2 = 10\alpha$,so $\alpha = 4t

Vedclass · Accepted Answer

more than $3$ but less than $6$

Vedclass · Answer

(D) The torque applied is $	au = rF = 2(20t - 5t^2) = 40t - 10t^2\, N\cdot m$.Using $	au = I\alpha$,we have $40t - 10t^2 = 10\alpha$,so $\alpha = 4t - t^2\, rad/s^2$.Integrating $\alpha$ with respect to time to find angular velocity $\omega$ (assuming $\omega_0 = 0$): $\omega = \int (4t - t^2) dt = 2t^2 - \frac{t^3}{3}$.The direction of motion reverses when $\omega = 0$,so $2t^2 - \frac{t^3}{3} = 0$,which gives $t^2(2 - \frac{t}{3}) = 0$. Thus,$t = 6\, s$.To find the angular displacement $	heta$,we integrate $\omega$: $	heta = \int_0^6 (2t^2 - \frac{t^3}{3}) dt = [\frac{2t^3}{3} - \frac{t^4}{12}]_0^6$.$	heta = \frac{2(216)}{3} - \frac{1296}{12} = 144 - 108 = 36\, rad$.The number of rotations is $N = \frac{	heta}{2\pi} = \frac{36}{2\pi} \approx \frac{36}{6.28} \approx 5.73$.Since $5.73$ is between $3$ and $6$,the correct option is $D$.

Similar Questions

The locus of the poles of focal chords of the parabola $y^2 = 4ax$ is

The heat evolved for the rise of water when one end of the capillary tube of radius $r$ is immersed vertically into water is (Assume surface tension $= T$ and density of water to be $\rho$)

Two solutions of $KNO_3$ and $CH_3COOH$ are prepared separately. The molarity of both is $0.1 \ M$ and their osmotic pressures are $P_1$ and $P_2$ respectively. The correct relationship between the osmotic pressures is:

Two infinite length wires carry currents $8A$ and $6A$ respectively and are placed along the $X$ and $Y$-axes. The magnetic field at a point $P(0, 0, d) \ m$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module