Question

A vertical wheel with a diameter of 39 cm starts from rest and rotates with a constant angular acceleration of 7.2 rad/s² around a fixed axis through its center counterclockwise.

a) Through what angle (in degrees) has the point initially at the bottom of the wheel traveled when t = 14 s? (Indicate the direction with the sign of your answer.)

b) What is the point's total linear acceleration at this instant? (Enter the magnitude in m/s².

Answer 1

Part A.

Using 2nd rotational kinematic equations:

= wi*t + (1/2)**t^2

= angular acceleration = 7.2 rad/sec^2

wi = Initial angular velocity = 0 rad/sec

t = time interval = 14 sec

So,

= 0*14 + (1/2)*7.2*14^2

= 705.6 rad = 705.6*180/pi deg = 40427.90 deg

= 40430 deg = rotation by bottom point

Part B.

total linear acceleration at this instant will be given by:

a = sqrt (at^2 + ac^2)

at = tangential acceleration = *R

R = radius of wheel = 39 cm/2 = 0.195 m

at = 7.2*0.195 = 1.404 m/s^2

ac = centripetal acceleration = w^2*R

w = final angular velocity at above given instant

Using 1st rotational kinematic equation:

w = w0 + *t = 0 + 7.2*14

w = 100.8 rad/sec

So,

ac = 100.8^2*0.195 = 1981.32 m/s^2

So,

a = sqrt (at^2 + ac^2) = sqrt (1.404^2 + 1981.32^2)

a = 1981.32 m/s^2

In three significant figures:

a = 1980 m/s^2

Let me know if you've any query.