Question

A stationary bicycle wheel of radiusis mounted in the vertical plane on a horizontal low-friction axle (Figur The 11.43).Thewheel has mass,$\mathit{M}$ all concentrated in the rim (the spokes have negligible mass). A lump of clay with mass $\mathit{m}$ falls and sticks to the outer edge of the wheel at the location shown. Just before the impact the clay has a speed $\mathit{v}$(a) Just before the impact, what is the angular momentum of the combined system of wheel plus clay about the center $\mathit{C}$ (b) Just after the impact, what is the angular momentum of the combined system of wheel plus clay about the center$\mathit{C}$in terms of the angular speed of the wheel? (c) Just after the impact, what are the magnitude and direction of the angular velocity of the wheel? (d) Qualitatively, what happens to the linear momentum of the combined system? Why?

Short answer

The rotational motion is counter-clockwise, the orbital angular momentum points out of the page.

Step-by-step solution

Definition of linear momentum.

The vector quantity linear momentum is defined as the product of an object's mass, m, and velocity, v.

On a horizontal low friction axle, a stationary bicycle wheel of mass$M$and radius$R$is attached in the vertical plane. A lump of clay with mass falls and clings to the wheel's outer edge at the area depicted in the diagram.

Fig. The figure shows the collision of the lump with the wheel.

Derive the expression for the magnitude of the angular momentum.

(a)Just before impact, the distance between the bicycle wheel's center and the clay is roughly equal to the radius of the bicycle, and the lump's component of velocity is$v\cos 45°$

The magnitude of the angular momentum is expressed as

$\left|L\right|=mvR\sin \theta$

Where,

$\theta$Angle between radius and velocity vector

$\nu$Velocity of the particle.

The radius vector is perpendicular to the component of the velocity vector, as seen in the diagram, which implies that$\theta =90°$

In this case, the angular moment of the wheel is zero and the parallel component of the velocity vector of the lump,$v_p=v\cos 45°$

Therefore, the magnitude of the angular momentum is

$\begin{array}{rcl}\left|L\right|& =& m{v}_{p}R\sin 90°\\ & =& mvR\cos 45°\\ & & \end{array}$………………..(1)

According to the right hand rule, if the rotating motion is counter-clockwise, the orbital angular momentum points out of the page.

Find the angular momentum of the system.

(b)The clay is attached to the bicycle wheel shortly after the contact. As a result, the distance between the bicycle wheel's centre and the clay is equal to the radius of the bicycle wheel.

The system's total mass is$M+m$

The relationship between the wheel's angular speed and the velocity of the clay is written as-

$\omega =\frac{v}{R}$

Angular momentum is expressed as-

$\left|L\right|=m{R}^2\omega$

Angular velocity of the clay is

$\left|L_c\right|=m{R}^2\omega$

Angular velocity of the wheel is

$\left|L_w\right|=m{R}^2\omega$

The system's angular momentum is equal to the sum of the clay and wheel's angular momentums.

As a result, the system's angular momentum is

$\begin{array}{rcl}\left|L\right|& =& \left|L_c\right|+\left|L_2\right|\\ & =& M{R}^2\omega +m{R}^2\omega \\ \left|L\right|& =& (M+m)R^2\omega \\ & & \end{array}$……………….(2)

(c)The overall angular momentum of the system is conserved, according to conservation of angular momentum. This means that the system's angular momentum is the same before and after the collision.

From the equations (1) and (2)

$mvR\cos 45°=(M+m)R^2\omega$

Therefore, the angular velocity of the wheel is

$\omega =\frac{mvR\cos 45°}{(M+m)R^2}$

According to the right hand rule, if the rational motion is counter-clockwise, the orbital angular momentum points out of the page.

(d)The clay's linear momentum reduces as the axle applies an upward impulsive force. The force exerted by the axle acts via no distance and does not work, yet the clay and striking wheel undergo a temperature rise.