Question

What happens to the initially stationary yo-yo in Fig. 11-25 if you pull it via its string with (a) force ${\overrightarrow{F}}_2$(the line of action passes through the point of contact on the table, as indicated), (b) force ${\overrightarrow{F}}_1$(the line of action passes above the point of contact), and (c) force ${\overrightarrow{F}}_3$(the line of action passes to the right of the point of contact)?

Short answer

(a) Force$\overrightarrow{F_2}$ will give 0 Nm torque.

(b) Force$\overrightarrow{F_1}$will give clockwise torque to yo-yo. So, it moves towards you.

(c) Force $\overrightarrow{F_3}$will give anticlockwise torque to yo-yo. So, it moves away from you.

Step-by-step solution

Step 1: Given

The forces acting on yo-yo are $\overrightarrow{F_1},\overrightarrow{F_2},\overrightarrow{F_3}$,

Determining the concept

The torque on the rotating object is a moment of force. It is equal to the cross product of the radius vector of rotation with the force acting on the rotating object.

Use the concept of torque to find the effects of the given forces.

Formulae are as follow:

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

Where, r is radius,F is force and $\tau$is torque

(a) Determining theforce F2→

Now, calculate torque due to force$\overrightarrow{F_2}$:

The pivot point is at ground so the angle between the pivot point and the force $\overrightarrow{F_2}$is${\theta }_2$

The torque will be,

$\left|\tau \right|=\left|{\overrightarrow{r}}_2\right|\times \left|\overrightarrow{F_2}\right|\text{sin}{\theta }_2$

The applied force is along the position vector so the angle is 0.

$\left|\tau \right|=\left|{\overrightarrow{r}}_2\right|\times \left|\overrightarrow{F_2}\right|\text{sin}\left(0^0\right)$

$\left|\tau \right|=0\text{Nm}$

Therefore, the magnitude of the torque is equal to zero.

(b) Determining the force F1→

Now, calculate torque due to force$\overrightarrow{F_1}$:

$\left|\tau \right|=\left|\overrightarrow{r_1}\right|\times \left|\overrightarrow{F_1}\right|\text{sin}{\theta }_1$

The angle is${90}^0$

$\left|\tau \right|=\left|\overrightarrow{r_1}\right|\times \left|\overrightarrow{F_1}\right|\text{sin}\left({90}^0\right)$

$\left|\tau \right|=\left|\overrightarrow{r_1}\right|\times \left|\overrightarrow{F_1}\right|$

From the right hand rule the direction of torque will be clockwise. So, it moves towards the right.

(c) Determining the force F3→

Now, calculate torque due to force$\overrightarrow{F_3}$ :

From the right hand rule the direction of torque will be anticlockwise. So, it moves towards the left.

Therefore, find the torques by using the concept of torque by considering the angles.

Take a pivot point where the yoyo is in contact with ground.