Question

A 14.0-kg solid homogeneous disk of radius 0.600 m is rotating with an initial angular speed of 50.0 rad/s about africtionless horizontal shaft. A 6.00-kg solid homogeneous disk of radius 0.400 m is initially at rest on the sameshaft. The disks are pushed into contact, as shown. Because of surface friction, the two disks eventually attain a finalcommon angular speed 64.(a) Find the final common angular speed of the system.(b) In the collision of the two disks, kinetic energy is not conserved because nonconservative (frictional) internalforces act during the contact. How much energy was lost to friction in the collision?

Answer 1

Given data

The mass of the first solid is m1 = 14 kg

The mass of the second solid is m2 = 6 kg

The radius of the first solid is r1 = 0.6 m

The radius of the second solid is r2 = 0.4 m

The angular speed of the first solid is w1 = 50 rad/s

The angular speed of the second solid is w2 = 0 rad/s

The expression for the final common angular speed of the system from the conservation of angular momentum is given as:

`\begin{gathered} I_1\omega_1+I_2\omega_2=(I_1+I_2)\omega_f \\ \omega_f=(I_1\omega_1+I_2\omega_2)/((I_1+I_2)) \\ \omega_f=\frac{(m_1(r_1)^2)/(2)\omega_1+(m_2(r_2)^2)/(2)_{}\omega_2}{((m_1(r_1)^2)/(2)+(m_2(r_2)^2)/(2))} \end{gathered}`

Substitute the value in the above equation.

`\begin{gathered} \omega_f=\frac{\frac{14\text{ kg}*(0.6m)^2}{2}*50\text{ rad/s +}\frac{6\text{ kg}*(0.4m)^2}{2}*0\text{ rad/s}}{\frac{14\text{ kg}*(0.6m)^2}{2}+\frac{6\text{ kg}*(0.4m)^2}{2}} \\ \omega_f=42\text{ rad/s} \end{gathered}`

Thus, the final common angular speed of the system is 42 rad/s.

(b)

The expression for the energy lost to friction in the collision is given as:

`\begin{gathered} \Delta E=E_i-E_f \\ \Delta E=(1)/(2)I_1(\omega_1)^2-(1)/(2)(I_1+I_2)(\omega_f)^2 \end{gathered}`

Substitute the value in the above equation.