Question

A uniform cylindrical turntable of radius 1.80 m and mass 27.2kg rotates counterclockwise in a horizontal plane with an initialangular speed of 4π rad/s. The fixedturntable bearing is frictionless. A lump of clay of mass2.39 kg and negligible size is droppedonto the turntable from a small distance above it and immediatelysticks to the turntable at a point 1.70 m to the east of the axis.

(a) Find the final angular speed of the clayand turntable.


What, if any, is the change in internal energy?
J


What, if any, is the amount of impulse imparted by the bearing?
kg· m/s

Answer 1

Radius of the cylindrical turn table is \(R=1.80 \mathrm{~m}\) Mass of the turn table is \(M=27.2 \mathrm{~kg}\) Initial angular speed of turn table is \(\omega_{i}=4 \pi \mathrm{rad} / \mathrm{s}\) Initial moment of inertia of turn table is \(I_{i}=\frac{1}{2} M R^{2}\)

$$ \begin{array}{l} =\frac{1}{2}(27.2 \mathrm{~kg})(1.70 \mathrm{~m})^{2} \\ I_{i}=39.04 \mathrm{~kg} \cdot \mathrm{m}^{2} \end{array} $$

Now a lump of clay of mass \(m=2.39 \mathrm{~kg}\) fall from small distance above it and immediately sticks to the turn table at a point \(r=1.70 \mathrm{~m}\) east of the axis Now the final moment of inertia is \(I_{f}=\frac{1}{2} M R^{2}+m r^{2}\)

$$ \begin{array}{l} =39.04 \mathrm{~kg} \cdot \mathrm{m}^{2}+(2.39 \mathrm{~kg})(1.70 \mathrm{~m})^{2} \\ =39.04 \mathrm{~kg} \mathrm{~m}^{2}+6.907 \mathrm{~kg} \mathrm{~m}^{2} \\ =45.947 \mathrm{~kg} \mathrm{~m}^{2} \end{array} $$

a) Let final angular speed be \(\omega_{f}\) If there are no external torques acting on the system then Angular momentum is conserved then \(I_{i} \omega_{i}=I_{f} \omega_{f}\) Then final angular speed of clayand turn table system is

$$ \begin{aligned} \omega_{f} &=\frac{I_{i} \omega_{i}}{I_{f}} \\ \omega_{f} &=\frac{\left(39.04 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(4 \pi \mathrm{rad} / \mathrm{s})}{45.947 \mathrm{~kg} \mathrm{~m}^{2}} \\ &=10.67 \mathrm{rad} / \mathrm{s} \end{aligned} $$

b) Initial mechanical energy is

$$ \begin{aligned} E_{i} &=\frac{1}{2} I_{i} \omega_{i}^{2} \\ &=\frac{1}{2}\left(39.04 \mathrm{~kg} \mathrm{~m}^{2}\right)(4 \pi \mathrm{rad} / \mathrm{s})^{2} \\ &=3079.35 \mathrm{~J} \end{aligned} $$

Final mechanical energy is \(E_{f}=\frac{1}{2} I_{f} \omega_{f}^{2}\)

$$ \begin{array}{l} =\frac{1}{2}\left(45.947 \mathrm{~kg} \mathrm{~m}^{2}\right)(10.67 \mathrm{rad} / \mathrm{s})^{2} \\ =2615.50 \mathrm{~J} \end{array} $$

Then mechanical energy is not conserved. Thus 464 Jof mechanical energ yis transformed into internal energy.

c) No, momentum is not conserved in this process. Because. Initial momentum is zero. Final momentum of the clay is \(m v=m v w_{f}\)

$$ \begin{array}{l} =(2.39 \mathrm{~kg})(1.70 \mathrm{~m})(10.67 \mathrm{rad} / \mathrm{s}) \\ =43.35 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} \end{array} $$

Thus turn table bearing imparts impulse of \(43.35 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\) along north into the turn table clay system and there after keeps changing the system momentum