Answers to Sample Test for Unit V

Potentially useful equations:

$\omega_{f}=\omega_{i}+\alpha (\Delta t)$ $v_{t}= r \omega$ $K_{\rm rot}=\frac{1}{2}I \omega^{2}$ $\sum F_{c} = F_{{\rm net},c} = m a_{c}$

$\Delta \theta = \omega_{i}(\Delta t) + \frac{1}{2}\alpha (\Delta t)^{2}$ $\vec{L} = I \vec{\omega}$ $W_{\rm net}=K_{f,\rm tot}-K_{i,\rm tot}$ $\sum F_{t} = m a_{t}$

$\tau = F_{\perp}r (= F r \sin \theta$) $\sum \vec{\tau} = I \vec{\alpha}$ $I= m r^{2}$ (point mass) $F_{G}= \frac{G m_{1}m_{2}}{r_{1,2}^{2}}$

$\omega_{f}^{2} = \omega_{i}^{2} + 2 \alpha (\Delta \theta)$ $a_{c}=\frac{v^{2}}{r}$ $K_{\rm lin}=\frac{1}{2}mv^{2}$ $\Delta \theta = \frac{1}{2}(\omega_{i}+\omega_{f}) \Delta t$

Longer Answers:

1. (a) Starting from $\omega_{f}=\omega_{i}+\alpha (\Delta t)$,
   
   $$\alpha = \frac{\omega_{f}-\omega_{i}}{\Delta t}$$
   
   
   $\alpha = 3$rad/s$^{2}$

   (b) With $\Delta t = 6 \rm s$, use $\Delta \theta = \omega_{i}(\Delta t) + \frac{1}{2}\alpha (\Delta t)^{2}$ to find that $\Delta \theta = 114$rad.

2. When rolling without slipping, the linear velocity and angular velocity are related by $v_{t}= r \omega$. The radius of the wheel is
   
   $$r= \rm\frac{27 in}{2}\times\frac{2.54\times 10^{-2} m}{1 in} = 0.343 m$$
   
   
   So $\omega = v_{t}/r= 43.7$rad/s ( $v_{t}=15\rm m/s$).

3. We need to have $a_{c}=9800\rm m/s^{2}$ and $r=0.09$m, and we know that $a_{c}=v^{2}/r = \omega^{2}r$ (from $v_{t}= r \omega$).
   
   $$a_{c} = \omega^{2}r$$
   
   
   
   
   $$\omega = \sqrt{\frac{a_{c}}{r}}$$
   
   
   $\omega = 330$rad/s = 3150 rev/min

4. (a) Consider the mass of the stick to be concentrated at the 50cm mark and balance the two torques: from the weight of the stick $\tau{\rm stick}$ (clockwise) and from the weight of the load $\tau_{L}$ (counter-clockwise). The torque equilibrium condition is $\sum \tau = 0$
   
   $$\tau_{\rm stick} - \tau_{L} = 0$$
   
   
   
   
   $$\tau_{\rm stick} = \tau_{L}$$
   
   
   
   
   $$\rm (0.27 kg)(9.8 m)(0.20 m)\sin 90^{\circ} = m(9.8 m)(0.30 m)\sin 90^{\circ}$$
   
   

   $m$ = 0.180kg or 180g

   (b) By balancing forces ( $F_{\rm net, y}= 0$), $F_{T}= m_{stick}g + m_{L}g =4.4$N.

5. (a) $I_{\rm tot}=I_{1}+I_{2}+I_{3}$, where $I_{1}=m_{1}r_{1}^{2}$, etc. $I_{\rm tot}=(11.25+1.28+8.67) {\rm kg m^{2}} = 21.2 {\rm kg m^{2}}$ (Note that the right-most mass is 1.7m from the axis!)

   (b) The torque is $\tau= (100 {\rm N})(1.5 {\rm m})\sin 30^{\circ}=75$Nm, so $\alpha = 3.5$rad/s$^{2}$ (from $\tau=I\alpha$)

6. $I_{i}= (2 {\rm kg})(2 {\rm m})^{2} + (2 {\rm kg})(2 {\rm m})^{2} = 16$kgm$^{2}$, and likewise find that $I_{f}=4$kgm$^{2}$. From conservation of momentum, $I_{f} \omega_{f} = I_{i} \omega_{i}$, and $\omega_{i}=5$m/s, so $\omega_{f}=20$rad/s.

7. See the answer to the handout problem (number 4 in set 3), which has identical procedure. The only real differences are the incline angle and length and the moment of inertia.
   
   $$v_{f}= \sqrt{\frac{4gh}{3} } =\rm\sqrt{\frac{4(9.8 m/s^{2})(0.97 m)}{3}} = 3.6 m/s$$