Question

A solid body rotates about a stationary axis accordig to the law theta=6t-2t^(3). Here theta, is in radian and t in seconds. Find (a). The mean values of thhe angular velocity and angular acceleration averaged over the time interval between t=0 and the complete stop. (b). The angular acceleration at the moment when the body stops. Hint: if y=y(t). then mean/average value of y between t_(1) and t_(2) is ltygt=(int_(t_(1))^(t_(2))y(t)dt))/(t_(2)-t_(1))

Answer 1

Given:

A solid body rotates about a stationary axis.

Ф =6t - 2t³

To Find:

(a) The mean values of the angular velocity and angular acceleration averaged over the time interval between t=0 and the complete stop.

(b) The angular acceleration at the moment when the body stops

Solution:

We know angular velocity,

• ω = dФ/dt
• ω = 6 - 6t²

Also angular acceleration is given by,

• α = dω/dt = -12t

At the moment the body stops rotating, its angular velocity = 0

Therefore,

• ω = 0
• 6 = 6t²
• t = 1 sec

(a) Mean value of angular velocity:

• ωm = $\frac{\int\limits^1_0 {w} \, dt }{1 - 0}$  = $\int\limits^1_0 {6 - 6t^{2} } \, dt$ = 6t - 2t³ = 6 - 2 = 4rad/s

   Mean value of angular acceleration:

• αm = $\frac{\int\limits^1_0 {\alpha } \, dt }{1 - 0}$ = $\int\limits^1_0 {-12t} \, dx$ = -6t = -6rad/s²

(b) The angular acceleration at the moment when the body stops,

• Value of α at t = 1sec
• α = 12rad/s²

The mean value of angular velocity is 4rad/s and mean value of angular acceleration is -6rad/s².

The angular acceleration at the moment when the body stops is -12rad/s²