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=-(dl)/(dt)`

Where l=Angular momentum

Using the formula

`\tau=(d(6.7))/(dt)=0`

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

`\tau=-(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^{(1)/(2)} kgm^2/s^{(3)/(2)}`

`\tau=-\frac{d(6.7t^{(1)/(2)})}{dt}=-6.7* (1)/(2)t^{-(1)/(2)}`

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

d.
`l=6.7t^(-2) kgm^2 s`

`\tau=-(d(6.7t^(-2)))/(dt)=13.4 t^(-3)`

`\tau=13.4(8.4)^(-3)=0.0226 k N m`