Question

A uniform disk with radius 0.320 m and mass 32.0 kg rotates in a horizontal plane on a frictionless vertical axle that passes through the center of the disk. The angle through which the disk has turned varies with time according to ?(t)=( 1.00 rad/s)t+( 6.10 rad/s2 )t2 .

a)What is the resultant linear acceleration of a point on the rim of the disk at the instant when the disk has turned through 0.300 rev ?

Express your answer with the appropriate units.

Answer 1

Given,

.....................(a)

Angular velocity is the first derivative of angular position:

Angular acceleration is the first derivative of angular velocity or the second derivative of angular position:

for,

Now,

Substitute this value in the Equation (a), We get:

After Solving the above Quadratic Equation, We get:

Now, Substitute above value in equation (1) we get:

Now, Centripetal Acceleration is:

Now, Tangential Acceleration is:

Now, The total linear acceleration is the vector sum of centripetal and tangential accelerations: