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) the angular speed of the drum

w = w(initial) +at

    = 0 + (0.40)(8)

    = 3.2 m/s

v = rw

  = 0.45(3.2)

  = 1.44

(ii) the centripetal acceleration of a point on the rim of the drum

ac = v²/r

     = 1.44²/0.45

     = 4.608 rad/s²

(iii) the tangential acceleration at that point

atan = r(angular acceleration)

         = 0.45(0.40)

         = 0.18 rad/s²

(iv) the resultant acceleration at that point

sqrt(tangential acceleration^2 + radial acceleration^2)

√4.608² + 0.18²

= 4.6 rad/s²