Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
PHY 116 From Newton to Einstein Model Answers to Exercise Sheet 5: Rigid Bodies Marks out of 30 1). A turntable has a kinetic energy of 0.0250 J when turning at 45.0 rev/min. What is the moment of inertia of the turntable about the axis of rotation? [1] [1] [1] [1] 2 E 0 . 05 2 = Answer: E = ½ Iω 2 ⇒ I = 2 = 0 . 00225 kg m ω (45 × 2π / 60)2 2). Small blocks, each with a mass m, are clamped at the ends and at the centre of a light rod of length L. Calculate the moment of Inertia of the system about an axis perpendicular to the rod and passing through a point ¼ of the length from one end. You can ignore the moment of inertia of the rod. [1] [3] [1] 2 2 2 2 2 3 L L 3 L 11L Answer: I = ∑ mi ri2 = m 2 + m 2 + m 2 = 16 4 4 4 i =1 3). A flywheel with a radius of 0.3m starts from rest and accelerates with a constant angular acceleration of 0.6 rad/s2. Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on the rim a) at the start; b) after it has turned through 60° and c) after it has turned through 120°. [1] [2] 2 2 Answer: aT = αr , a R = rω = r (αt ) = 2rαθ (a) At t = 0, θ = 0, aT = 0.6 × 0.3 = 0.18 m s −2 aR = 0 (b) [1] [1] a = aT = 0.18 m s −2 At θ = 60°, aT = 0.6 × 0.3 = 0.18 m s −2 a R = 2 × 0.3 × 0.6 × π / 3 = 0.377 m s −2 (c) [1] a = aT2 + a R2 = 0.418 m s −2 At θ = 120°, aT = 0.6 × 0.3 = 0.18 m s −2 a R = 2 × 0.3 × 0.6 × 2π / 3 = 0.754 m s −2 a = aT2 + a R2 = 0.775 m s −2 [1] 4). A cord is wrapped around the rim of a wheel 0.25m in radius, and a steady pull of 40.0 N is exerted on the cord. The wheel is mounted on a frictionless bearing on a horizontal shaft through its centre. The moment of inertia of the wheel about the shaft is 5.00 kg.m2. Compute the angular acceleration of the wheel. Answer: [2] Torque τ = Fr = 10 N m. τ 10 Also τ = Iα, so α = = = 2 rad s −2 [3] 5 I 5). Under some circumstances a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of the neutron start is roughly 1014 times as great as that of ordinary solid matter. Suppose we represent the start as a uniform solid rigid sphere, both before and after collapse. The star’s initial radius was 7.0 x 105 km (comparable to the Sun); its final radius is 16 km. If the original star rotated once in 30 days, find the angular speed of the neutron star. Answer: Angular momentum L is conserved, and L = Iω = c MR2ω where c is some factor appropriate to the sphere. So [1] cMR12ω1 = cMR22ω 2 [2] 2 [1] 5 2 7 × 10 R 2π ω 2 = ω1 12 = = 4639 rad s -1 R2 16 30 × 24 × 60 × 60 6). Occasionally, a rotating neutron star undergoes a sudden speedup called a glitch. One explanation is that a glitch occurs when the crust of the neutron start settles slightly, decreasing the moment of inertia about the rotation axis. A neutron star with angular speed ω0 = 70.4 rad/s underwent such a glitch in October 1975 that increased its angular speed to ω = ω0 + Δω, where Δω/ω0 = 2.01 x 10-6. If the radius of the neutron star before the glitch was 11 km, by how much did the radius decrease in the starquake? Assume that the neutron star is a uniform sphere. Answer: Using cMR12 ω1 = cMR22 ω 2 R22 = R12 ω1 ω2 [1] [1] [1] ∆ω0 ω0 ∆ω0 ω1 = R1 = R1 1 − = R1 1 − R2 = R1 ω2 ω0 + ∆ω0 ω0 2ω0 ∆ω0 [1] ∆R = R1 − R2 = R1 = 11 km × ½ × 2.01 × 10 −6 = 11 mm [1] 2ω 0