Question

String is wrapped around an object of mass 1.5kg and moment of inertia 0.0015kg-${\mathbf{m}}^{\mathbf{2}}$(the density of the object is not uniform). With your hand you pull the string straight up with some constant force F such that the center of the object does not move up or down, but the object spins faster and faster (Figure 9.62). This is like a yo-yo; nothing but the vertical string touches the object. When your hand is a height ${\mathit{y}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{m}$above the floor, the object has an angular speed ${\mathbf{\omega }}_{\mathbf{0}}\mathbf{=}\mathbf{12}\mathbf{rad}\mathbf{/}\mathbf{s}$. When your hand has risen to a height y=0.35m above the floor, what is the angular speed of the object? Your answer must be numeric and not contain the symbol F.

Short answer

The angular speed of the object is $41037\mathrm{rad}/\mathrm{s}$.

Step-by-step solution

Identification of given data

The given data is listed below as,

• The moment of inertia of the string is,$I=0.0015\mathrm{kg}-{\mathrm{m}}^2$
• The mass of the object is,m=1.5kg
• Initially, the height of the hand above the floor is,$y_0=0.25\mathrm{m}$
• The initial angular speed of the object is, ${\omega }_0=12\mathrm{rad}/\mathrm{s}$
• Finally, the height of the hand above the floor is, y=0.35 m

Significance of the angular speed

Angular speed is described as the ratio of the change in the angular rotation to the time. However, it is also known as rotational or angular velocity. This is also helps in understanding the rotation of an object.

Calculation of the work done

The equation of the change in energy of the system is:

$W=K_{trans}+K_{rot}$ …(i)

Here, $K_{tras}$is transitional kinetic energy is zero because mass is not moving and $K_{rot}$is rotational kinetic energy.

Equation (i) can be expressed as follows:

$k_{rot}=W$ …(ii)

The equation of the force can be calculated as:

$F=m.g$

Here, m is the mass of the object, and g is the acceleration due to gravity.

The equation of the work done is expressed as:

$W=F\left(y-y_0\right)$

Here, $F$is the force exerted, y is the final height and $y^0$is the initial height.

Substitute all the values in above equation.

$W=m.g.\left(y-y_0\right)$

Determination of the angular speed for the system

As the moment of inertia is given, the equation of the rotational kinetic energy becomes:

$K_{rot}=\frac{1}{2}I{\omega }^2-\frac{1}{2}I{{\omega }^2}_0$

Here, I is the moment of inertia, $\omega$is the final angular speed, and ${\omega }^2$is the initial angular speed.

Substitute all the values in equation (ii).

$$\begin{gathered} \frac{1}{2}I{\omega }^2-\frac{1}{2}I{{\omega }^2}_0=m.g.\left(y-y_0\right) \\ I\left({\omega }^2-{{\omega }^2}_0\right)=2.m.g.\left(y-y_0\right) \\ \left({\omega }^2-{{\omega }^2}_0\right)=\frac{2.m.g.\left(y-y_0\right)}{I} \\ \omega =\sqrt{{{\omega }_0}^2+\frac{2.m.g.\left(y-y_0\right)}{I}} \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} \omega =\sqrt{{\left(12\mathrm{rad}/\mathrm{s}\right)}^2+\frac{2\times \left(1.2\mathrm{kg}\right)\times \left(0.35\mathrm{m}-0.25\mathrm{m}\right)}{0.0015\mathrm{kg}.{\mathrm{m}}^2}} \\ =\sqrt{144{\mathrm{rad}}^2/{\mathrm{s}}^2+\frac{2\times \left(1.2\mathrm{kg}\right)\times \left(0.98{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{0.0015\mathrm{kg}.{\mathrm{m}}^2}} \\ =\sqrt{144{\mathrm{rad}}^2/{\mathrm{s}}^2+\frac{\left(23.52\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{0.0015\mathrm{kg}.{\mathrm{m}}^2}} \\ =144\mathrm{rad}/s \end{gathered}$$

Thus, the angular speed of the object is $144\mathrm{rad}/s$.