Question

A cylindrical drum has a radius of 0.45 m and is initially at rest. It is then given an angular acceleration of 0.40 rad s−2. At time t = 8.0 s calculate (i) the angular speed of the drum, (ii) the centripetal acceleration of a point on the rim of the drum, (iii) the tangential acceleration at that point, and (iv) the resultant acceleration at that point.

Answer 1

(i)

$$\omega = \alpha t=(0.4)(8)=3.2 rad/s$$

(ii)

$$a_c=\omega^2r=(3.2)^2 0.45=4.6 m/s^2$$

(iii)

$$a_t=\alpha r=(0.4)(0.45)=0.18 m/s^2$$

(iv)

$$a= \sqrt{0.18^2+4.6^2} =4.6 m/s^2$$