PHS1022-2_2003_Solutions

PHS1022-2_2003_Solutions - Monash University Semester Two...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Monash University Semester Two Examinations 2003 Faculty of Science - School of Physics and Materials Engineering PHS1022 Paper 2 Physics - S O L U T I O N S Students are strongly advised to attempt questions before looking at these solutions . SECTION A - ROTATIONAL MECHANICS AND GRAVITY QUESTION A1 (2+2+3 =7 marks) A flywheel with rotational inertia I = 30 kg m 2 is being slowed down, with its angular speed given by rad/s. (a) What is the mean angular speed and the angular displacement in radian from t = 0 to t = 20 s? ω (20) = 200 - (2)(20) = 160 rad/s. So (b) What is the magnitude of the braking torque? (Magnitude = 60 Nm.) (c) What is the power dissipated from the flywheel at t = 0? P = τω = (60 Nm)(200 rad/s) = 12000 W. (Analogous to P = Fv). Alternatively, find kinetic energy K(t) at t = 0 and 20 s, then work done = K (20) - K (0), then power = work/time. QUESTION A2 (5 + 3 = 8 marks) A uniform solid cylinder and a hoop both have the same outer radius R and mass M. (The hoop has its mass uniformly distributed in a circle of radius R). They are allowed to roll from rest down an inclined plane to determine if they reach the bottom at the same time with the same speed. (a) Write expressions for the energy of the objects at the bottom of the plane and hence give a clearly reasoned prediction for the outcome of this experiment. Cylinder : Hoop: Since potential energy mgh is converted to K , both have the same K at the start/bottom of the slope. Since the cylinder’s rotational inertia is smaller than the hoop’s, then the cylinder will have larger angular speed than the hoop. (alternatively, looking at the two expressions for K , for the same value of K , the cylinder has the largest ω . (b) Determine the speed of the hoop if it had descended through a vertical height h . Since K = MR 2 ω 2 = mgh , then for hoop:
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
QUESTION A3 ( 4 + 4 = 8 marks) a) A car travels at uniform velocity past an observer O, as shown. Does its angular momentum about O vary as the car moves from the closest point to further away? Give a reason. L is constant , since; resultant torque on car = 0; alternatively, | L| = m r v sin θ = mvd = constant b) An asteroid makes a close approach to Jupiter, so its trajectory changes as shown. Does its angular momentum about Jupiter vary? Explain carefully. The force between asteroid and planet is along the line of the planet. Torque on asteroid = r × F = 0 since the two vectors are collinear, with θ = 180 E . So the angular momentum is unchanged, since rate of change of ang mom = net torque. QUESTION A4 (2 + 2 + 3 = 7 marks) A spinning bicycle wheel (angular speed ω w , rotational inertia I w ) is held by a person who is initially at rest standing on a platform which is free to rotate. (As viewed, the front of the wheel is moving to the right.) The person then turns the bicycle wheel through 180 so its axis is vertical again. (a)
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 14

PHS1022-2_2003_Solutions - Monash University Semester Two...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online