Question

The road wheel of a motor vehicle increases speed uniformly from 50 rev/min to 1100 rev/min in 40 seconds.calculate the angular acceleration of the wheel in rad/seconds squared,if the diameter of the wheel is 0.7 meters what is the linear acceleration of a point on a Tyre trend.

Answer 1

angular acceleration= change in anglevelocity/time=(1100-50)2PI/60)/40 rad/sec

linear tangential acceleration= angularAccelerationabove*radius

Answer 2

Vo = 50rev/min * 1min/60s * 6.28rad/rev = 5.23 rad/s.
V = 1100rev/min * 1min/60s * 6.28rad/rev = 115.1 rad/s.

a. V = Vo + a*t = 115.1.
5.23 + a*40 = 115.1,
a = 2.75 rad/s^2.

b. Circumference = pi * Dia. = 3.14 * 0.7 = 2.2 m.
Linear acceleration = 2.75rad/s^2 * 1rev/6.28rad * 2.2m/rev =

Answer 3

What is the answer of the linear acceleration