Rotational motion problems

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Pierce Cunningham
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1 Rotational motion problems. (Massive pulley) Masses m and m 2 are connected by a string that runs over a pulley of radius R and moment of inertia I. Find the acceleration of the two masses, as well as the tension in each string segment. Write down equations for the net force acting on each mass, as well as the net torque acting on the pulley (making sure to be consistent about which direction of motion is positive, including the sense of rotation of the pulley): F = m g sin θ + T = m a, F2 = m 2 g T 2 = m 2 a 2, τ = RT2 RT = Iα. If there is enough friction between the string and the pulley, then whenever a length of string x passes over the pulley, the pulley will rotate by an angle θ = x/r. This implies αr = a = a 2. Solving the above equations then yields a = a = αr = m g sin θ + m 2 g m + m 2 + I/R 2, T = m 2 (m 2 + I/R 2 ) sin θ m + m 2 + I/R 2 m g, T 2 = (m + I/R 2 ) m sin θ m + m 2 + I/R 2 m 2 g.

2 2. (Non-symmetric rotation) A massless rod of length 2l is attached to a vertical axle rotating at angular velocity ω. The rod is tilted at an angle α relative to the horizontal, and has small masses m attached at each end. Find the angular momentum vector L. Is it parallel to the angular velocity vector ω? Why not? This example is analyzed in considerable detail in K&K, starting on page 292. The short answer is that L is directed perpendicular to the rod, and has magnitude L = 2ml 2 ω cos α. It is not parallel to ω, because the object is not symmetric about the vertical axis; instead they are related by L = Ĩ ω, where Ĩ is the inertia tensor and the denotes matrix multiplication. L rotates as the object turns, which implies the existence of a nonzero torque applied to the rotating axle by the bearings. 3. (SHM via energy) One way to identify simple harmonic motion is by an equation of motion: if x is some dynamical quantity that satisfies an equation of the form ẍ + ω 2 x = 0, then x will undergo SHM with angular frequency ω [i.e. x(t) = A cos(ωt + φ 0 )]. An alternative way to identify SHM is by looking at the total energy of a mechanical system. If the total energy depends on some quantity x and its time derivative ẋ in the following way, E = 2 m(ẋ2 + ω 2 x 2 ) + constant, then x will undergo SHM at angular frequency ω. Use this energy approach to show that for a simple pendulum undergoing small oscillations, ω = g/l. 2

3 When the pendulum is displaced by a small angle θ from its equilibrium position (which is straight down), the gravitational potential energy is U = mgl( cos θ). For small θ, cos θ θ2 2, so U 2 mglθ2. Thus the total energy of the simple pendulum is E = K + U = 2 ml2 θ2 + 2 mglθ2 = 2 ml2 ( θ2 + g l θ2). This is of the required form for SHM, with ω 2 = g/l. 4. (K&K Problem 6.7) A rod of length l and mass m, pivoted at one end, is held by a spring at its midpoint and a spring at its far end, both pulling in opposite directions. The springs have spring constant k, and at equilibrium their pull is perpendicular to the rod. Find the frequency of small oscillations about the equilibrium position. Either compute the torque from the two springs and gravity, and show it is proportional to sin θ θ for small angles, or compute the potential energy of the system and show it goes like θ 2. Either way, you should find ω 2 = 5 k 4 m 3 g 2 l Note that if 5kl < 2mg then the rod will not oscillate when displaced slightly from equilibrium, but will instead just fall over. 5. (K&K Problem 6.4) A spaceship is sent to investigate a planet of mass M and radius R. While hanging motionless in space at a distance 5R from the center of the planet, the ship fires an instrument package with speed v 0, as shown in the sketch. The package has mass m, which is much smaller than the mass of the spaceship. For what angle θ will the package just graze the surface of the planet? 3

4 Use conservation of angular momentum (about the center of the planet, which is presumed stationary) combined with conservation of energy. 5Rmv 0 sin θ = Rmv 2 mv2 0 GMm 5R = 2 mv2 GMm R = sin θ = + 8 GM 5 5 v0 2R Note that if v 0 is very small (specifically if v0 2 < GM/5R), then there is no way to make the package graze the planet s surface: gravity will pull the package towards the planet until they collide. On the other hand, if v 0 is very large (let s say v 0 = ), then the package will travel in a straight line, and geometry tells us that we want sin θ = /5, in agreement with our answer. 6. (Rolling with slipping) A disk with mass m, radius r, and moment of inertia I is given a push so that its initial horizontal velocity is v 0, but its initial angular velocity ω 0 is zero. It moves along a horizontal surface with coefficient of kinetic friction µ. Find the linear velocity v(t) and angular velocity ω(t) of the disk as functions of time. The definition of rolling without slipping is that the element of the disk that is in contact with the surface at any given instant has, at that instant, zero linear velocity. That is, the rotational motion from the rolling of the object is offset by the linear motion of the object s center of mass. This translates into the rolling without slipping condition v CM = ωr. For the current problem, when the disk starts out it is not rolling without slipping, since v 0 ω 0 r. This means that the element of the disk that is in contact with the surface is sliding along the surface. Thus, there is a (kinetic) friction force acting on the disk. To see which direction this frictional force acts, look closely at the element of the disk in contact with the surface. At t = 0 it has linear velocity v 0 relative to the surface; in general at time t this will be v(t) ω(t)r. So long as this quantity is positive, the element in question is moving to the right, hence friction will act to the left. This force will produce a torque on the disk, causing it to start rolling. 4

5 Now that we understand the motion qualitatively, we can solve for the desired quantities. The friction force is f = µn = µmg, acting to the left. This is the only force acting horizontally, so ma = F x = f = µmg = a = µg. This acceleration is constant, so v(t) = v 0 µgt. The frictional force also generates a torque about the center of mass of the disk, τ = rf. This torque acts in the clockwise direction (which we shall call the positive sense of rotation), so and ω(t) = (µmgr/i)t. Iα = τ = rf = µmgr = α = µmgr/i, Does this motion (decreasing linear velocity and increasing angular velocity) continue forever? Obviously not. This solution depends on the assumption that there is a frictional force acting on the disk. But once the linear and angular velocities have self-corrected to the point where v(t) = ω(t)r, then the disk will be rolling without slipping, and no friction force will act on it. The time t at which we achieve the rolling without slipping condition is v(t ) = ω(t )r = v 0 µgt = (µmgr 2 /I)t = t = After this time, v(t) = v and ω(t) = ω are constant, where v = v 0 µgt = + I/mr 2 v 0, ω = (µmgr/i)t = Graphically, the functions v(t) and ω(t) look like this: v 0 µg( + mr 2 /I). + I/mr 2 v 0 r. 5

Exam 3 Practice Solutions

Exam 3 Practice Solutions Multiple Choice 1. A thin hoop, a solid disk, and a solid sphere, each with the same mass and radius, are at rest at the top of an inclined plane. If all three are released at