Question

A cyclist accelerates from rest at a rate of \({\bf{1}}{\bf{.00}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). How fast will a point at the top of the rim of the tire (diameter = 0.80 cm) be moving after 2.25 s? [Hint: At any moment, the lowest point on the tire is in contact with the ground and is at rest — sees Fig. 8–57.]

FIGURE 8-57 Problem 79

Short answer

The point at the top of the tire's rim will be moving at a speed of 4.50 m/s after 2.25 s.

Step-by-step solution

Relation between linear and angular velocity

If an object is rotating about a fixed axis, thelinear velocity (v) of any point on the object located at a distance r from the axis of rotation is related to the angular velocity\(\left( \omega \right)\)of the objectby the following relation:

\(v = r\omega \)

In this problem,the linear velocity of the axle of thetire and angular velocity of the tire are related to each other by the above relation.

Given information

The acceleration of the tire is\(a = 1.00\;{\rm{m/}}{{\rm{s}}^2}\).

The diameter of the tire is d = 0.80 cm.

The initial linear speed of the axle of the tire relative to the ground is\({v_0} = 0\;{\rm{m/}}{{\rm{s}}^2}\).

Time taken,\(t = 2.25\;{\rm{s}}\)

Consider thedirection of motion of the tire, i.e., the right direction to be positive.

Determination of the speed of axle of the tire

Let the linear speed of the axle of the tire relative to the ground after time t be v.

Using the kinematic equation of motion for the motion of the axle of the tire relative to the ground, you will get:

\(\begin{aligned}{c}v = {v_0} + at\\ = \left( {0\;{\rm{m/s}}} \right) + \left( {1.00\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {2.25\;{\rm{s}}} \right)\\ = 2.25\;{\rm{m/s}}\end{aligned}\) ... (i)

Thus, the velocity of the axle relative to the ground in the right direction is

\({\vec v_{{\rm{ag}}}} = v = 2.25\;{\rm{m/s}}\)to the right of the tire.

Determination of the speed of the top of the tire

At any instant, the lowest point of the tire in contact with the ground is at rest. Therefore, the velocity of the lowest point will be zero.

If the axle of the tire is taken as the reference frame and \(\omega \) as the angular velocity of the tire, then every point on the outer circumference of the tire will be moving with the speed:

\(v = r\omega \)

Thus, you can say that the velocity of the point at the top of the tire relative to the axle of the tire is:

\({\vec v_{{\rm{ta}}}} = v\)

The direction of this velocity is along the tangent at the top point, i.e., along the right of the tire.

So, the velocity of the top of the tire relative to the ground is:

\(\begin{aligned}{c}{{\vec v}_{{\rm{tg}}}} = {{\vec v}_{{\rm{ta}}}} + {{\vec v}_{{\rm{ag}}}}\\ = v + v\\ = 2v\end{aligned}\)

The direction of this velocity is toward the right.

Thus, using (i), the velocity of the top of the tire relative to the ground is:

\(\begin{aligned}{c}{{\vec v}_{tg}} = 2v\\ = 2\left( {2.25\;{\rm{m/s}}} \right)\\ = 4.50\;{\rm{m/s}}\end{aligned}\)