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

Given that,

radius = 0.45 m.

angular acceleration = 0.4 rad/s^2 .

time = 8 s.

To find:

1). Angular speed of drum,

We know that angular speed of drum is: ω=αt

     ω=0.4 × 8

     ∴ω=3.2 rad/s.

2). Centripetal acceleration,

We know that centripetal acceleration is:  ac=rω^2

ac=rω^2

ac=0.45×3.2^2

∴ac=4.608m/s^2.

3) Tangential acceleration,

We know that tangential acceleration is: at=rα

at=rα

at=0.45×0.4

∴at=0.18m/s^2.

4) Resultant acceleration,

We know that resultant acceleration is: a=√(ac^2+at^2)

a=√(ac^2+at^2)

a=√(4.608^2+0.18^2)

a= 4.611m/s^2