Question

A passenger compartment of a rotating amusement park ride contains a bench on which a book of mass

mb is
placed, as indicated by the dot in the left figure above. The compartment moves with a constant angular speed
about the center of the ride along a circular path of radius R. The bench remains horizontal throughout the
compartment’s motion. The right figure above shows a magnified view of the compartment.
The graph below shows the horizontal ( ) x component of the book’s position as a function of time, where the
+x-direction is to the right.
(a)
i. Determine the period of revolution of the book.
ii. Calculate the tangential speed vb (not the angular speed) of the book.

Answer 1

a) 120 s

b) v = 0.052R [m/s]

Step-by-step explanation:

a)

The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).

The graph of the problem is missing, find it in attachment.

To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.

The first point we take is t = 0, when the position of the book is x = 0.

Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.

Therefore, the period is

T = 120 s - 0 s = 120 s

b)

The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.

The perimeter of the wheel is:

`L=2\pi R`

where R is the radius of the wheel.

The period of revolution is:

`T=120 s`

Therefore, the tangential speed of the book is:

`v=(L)/(T)=(2\pi R)/(120)=0.052R`