PHYS*1080 Sample Quiz 7

  1. A stone of mass \(200\; \text{g}\) is swung on a horizontal circle of radius \(1 \; \text m\) with a speed of \(5\; \text{m/s}.\) (4 marks)

    (a) What is the tension in the string?

    (b) How much work does the tension force do in 5 revolutions?

    (c) Suppose it had taken \(10 \; \text s\) to get this speed from rest. What then is the angular acceleration?
    Part Solution Marks
    (a) \(T = F = mv^2/r = 0.2 \; \text{kg} \times (5 \; \text{m/s}^2/1\; \text m = 5\; \text N)\)  (1) for  \(T = \frac{mv^2}{r} \)
    (0.5) Answer
    (b) Work done by tension = 0 since there is no displacement in the direction of the force.  (1)
    (c) \(\omega_o = 0 \quad \quad \omega = v/r = 5/1\; \text s ^{-1} \\ t = 10\; \text s \quad \; \alpha = ? \\ \alpha = (\omega-\omega_o)/t = (5-0)/10 = 0.5\; \text{rad/s}^2\)

    (1) \(\alpha = \frac{\omega - \omega_o}{t}\)

    (1.5) Answer



     
  2. A figure skater rotating at a constant 1.5 revolutions per second has a rotational kinetic energy of \(1500 \; \text J\). (4 marks)

    (a) What is her moment of inertia?

    (b) What net torque (moment) is acting on the skater? 

    (c) What is her angular momentum?
    Part Solution Marks
    (a) \(K.E. = \frac{1}{2}I \omega^2 \quad \quad \therefore I = 2(K.E.)/\omega^2 \\ \therefore I = (2 \times 1500 \; \text J)/(1.5 \times 2\pi \;\text{rad/s}^2 = 33.8\; \text {kg m}^2)\) (1.0)
    (0.5)
    (b) \(\sum \tau = l \alpha \quad \quad \alpha = 0 \\ \therefore\; \text{Net torque}\; = 0\) (1.0)
    (c)

    \(\begin {align} \text{Angular momentum} & = I\omega \\ & = 34\; \text{kg m}^2 \times 3 \pi\; \text{rad/s}\\ & = 3.2 \times 10^2 \; \text{kg m}^2 /\text s \end {align}\)

    \((\text{or} I\omega = 2 (K.E.)/ \omega)\)

    (1.0)



    (0.5)


     
  3. A molecule of mass \(1.0 \times 10^{-25} \; \text {kg}\) is traveling at a constant speed of \(2.0 \times 10^{-6}\; \text{m/s}\) in a circle of radius \(3.0 \times 10^{-8}\text m\). What is the force (magnitude and direction) on the molecule? (1.5 marks)
    Solution Marks
    \(F = \frac{mv^2}{r} = \frac{1.0 \times 10^{-25}\; \text{kg} \times (2.0 \times 10^{-6}\; \text{m/s})^2}{3.0 \times 10^{-8}\text m}\) (0.5)
    \(= 1.3 \times 10^{-29}\; \text N\) (0.5)
    Towards the centre of rotation (0.5)


     
  4. The earth exerts a gravitational force \(2 \times 10^{20}\) on the moon, which travels a distance \(2.4 \times 10^9\) along the complete circumference of its circular orbit around the earth in  a time of \(2.4 \times 10^6 \; \text s\) (about 28 days). How much work does the earth do on the moon in each orbit? (0.5 marks)
    Solution Marks
    None: the force is at right angles to the motion (0.5)