Question

A particle is to move in an xy plane, clockwise around the origin as seen from the positive side of the z axis. In unit-vector notation, what torque acts on the particle at time t = 8.4 s if the magnitude of its angular momentum about the origin is (a)6.7 kg·m2/s, (b)6.7t2 kg·m2/s3, (c)6.7t1/2 kg·m2/s3/2, and (d)6.7/t2 kg·m2*s?

Answer 1

Answer with Explanation:

We are given that

Time, t=8.4 s

a.Angular momentum=$6.7 kg m^2/s$

We know that torque acts on the particle

$\tau=-\frac{dl}{dt}$

Where l=Angular momentum

Using the formula

$\tau=\frac{d(6.7)}{dt}=0$

b.$l=6.7 t^2kg m^2/s^3$

$\tau=-\frac{d(6.7t^2)}{dt}=-13.4t$

Substitute t=8.4

$\tau=-13.4(8.4)=-112.56 k N-m$

c.$l=6.7t^{\frac{1}{2}} kgm^2/s^{\frac{3}{2}}$

$\tau=-\frac{d(6.7t^{\frac{1}{2}})}{dt}=-6.7\times \frac{1}{2}t^{-\frac{1}{2}}$

$\tau=-\frac{6.7}{2}(8.4)^{-\frac{1}{2}}=-1.155k N-m$

d.$l=6.7t^{-2} kgm^2 s$

$\tau=-\frac{d(6.7t^{-2})}{dt}=13.4 t^{-3}$

$\tau=13.4(8.4)^{-3}=0.0226 k N m$