acceleration

Question

2.2. a pulley, with an effective diameter of 16cm, is to drive a flat belt. The pulley is to be uniformly accelerated from rest to 5400 r/min during 540 revolutions. Calculate the ffg: 2.2.1. the uniform angular acceleration of the pulley in rad/s^2 2.2.2. the linear acceleration of a point on the belt in m/s^2.

Answer 1

(2.2.2)

Diameter of the pulley = 16 cm.

Now we know that .

Initial Velocity = 0

α = 565.5 / 6

α = 94.24 rad/sec²

We know that relationship between Linear accerlation and angular accerlation is

Where a = Linear accerlation

α = Angular accerlation

r = radius.

a = 8 * 10^-2 * 94.24

a = 7.53 m/sec²

Linear Accerlation = 7.53 m/sec².

Answer 2

(2.2.1)

The diameter of the pulley = 16 cm.

Radius of the pulley = 16/2 = 8  cm.

Angular speed = 5400 r/min

Rotations = 540 rotations.

Angular Speed  = rotations/time

time = Rotations / Angular Speed

t = 540 / 5400

t = 0.1 min * 60 sec

t = 6 sec.

We know that angular velocity = 5400 rpm

Angular Velocity = 5400 rev/min * 2π rad/rev * 1min/60sec = 565.4 rad/sec.

Angular Velocity = 565.4 rad/sec.

Now we know that .

Initial Velocity = 0

α = 565.5 / 6

α = 94.24 rad/sec²

Angular Accerlation = 94.24 rad/sec².