# No Brain Too Small PHYSICS. 2 kg

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1 MECHANICS: ANGULAR MECHANICS QUESTIONS ROTATIONAL MOTION (2014;1) Universal gravitational constant = N m 2 kg 2 (a) The radius of the Sun is m. The equator of the Sun rotates at a rate of 14.7 degrees per day. (i) (ii) Show that the period of rotation of a particle located on the equator of the Sun is s. Calculate the linear speed of a particle at the Sun s equator. Gravity may cause the rotating inner core of the Sun to collapse down to a much smaller radius. Explain how this will affect the angular speed of the inner core. (c) The mass of Mercury is kg. Mercury has a period of rotation of s. Show that a satellite needs to be positioned m from the centre of Mercury so that it remains stationary from the point-of-view of an observer on that planet. A space probe spins around an axis, as shown. An instrument comes loose from the space probe. Explain why this loss of mass will have no effect on the angular speed of the space probe. TOSSING BALLS (2013;1) A hollow ball has a mass of kg and radius m. The ball is thrown vertically upwards from rest. It rises through a height of 1.40 m then drops down again. When it is released, it is moving upwards at 5.24 m s -1 and rotating at 2.70 revolutions per second. During the throwing action, a tangential force of N is applied to the surface of the ball for a period of s. (a) (c) Show that the angular speed of the ball when it is released is 17.0 rad s 1. Show that the average angular acceleration of the ball before it is released is 67.9 rad s 2. Calculate the rotational inertia of the ball. For the following two situations, explain whether the height to which the ball rises will be less than, greater than, or the same as 1.40 m. Ignore the effects of air resistance. (i) (ii) The ball is not rotating, but is given the same linear speed when it is released. The ball is solid instead of hollow, but has the same mass and radius. The same amount of total work is done to give the ball its linear and rotational motion, and it has the same angular speed.

3 ROTATIONAL MOTION (2010;1) The bicycle wheel shown in the diagram consists of a set of lightweight spokes that radiate out from the hub to the rim. The wheel has a rotational inertia of kgm 2. The rim diameter of the wheel is m and the tyre diameter is m. (a) (i) The wheel rotates at a steady speed. It completes one revolution in s. Calculate its angular velocity. (ii) Calculate the rotational kinetic energy of the wheel. The rotational inertia of a hollow cylinder of mass m, and radius r, can be calculated from I = mr 2. Discuss how you could use this formula to calculate the rotational inertia of the wheel and what assumptions you would have to make. (c) A cyclist is riding a bike at a speed of 5.50 ms 1. Both wheels on the bicycle are the same as the one in the diagram. The bicycle and its rider have a total mass 68.4 kg. Calculate the work that was done to accelerate to this speed. You may ignore any frictional effects. Two common types of bicycle are road bikes and mountain bikes. Mountain bike wheels generally have a lower rotational inertia than road bike wheels. A mountain bike wheel and a road bike wheel are both accelerated at the same rate. Explain why it is easier to accelerate the mountain bike wheel than the road bike wheel. QUESTION TWO (2009;2) To start the plane, Sam has to turn the propeller by hand. When the propeller reaches a high enough angular velocity, the engine will be able to start. The propeller has a rotational inertia of 16.5 kgm 2. The engine will start when the propeller is turning at 100 revolutions per minute (rpm). (a) Show that 100 rpm is 10.5 rads -1 Sam is able to apply a constant torque to the propeller for 1/3rd of a revolution. The propeller is at rest when Sam starts to turn it. Calculate the minimum constant torque he must apply to start the engine. Sam wants to fit a new propeller, of the same length, which will be easier to accelerate to 100 rpm. Explain how the new propeller might be different to the existing propeller. When the engine is shut off, the propeller takes 20.0 s to come to a complete stop from an angular velocity of 37.7 rads -1. Calculate the number of turns the propeller completes while it slows to a stop. State any assumption(s) you have made.

4 ROTATIONAL MOTION (2008;1) Revolving doors are used in many big buildings. You may assume that the effects of friction can be ignored in this question. Jenny enters a revolving door, which is initially stationary (above diagram). She pushes on the door at point A: it accelerates at 0.48 rads -2. She stops pushing when it reaches an angular velocity of 0.58 rads -1. (a) Jenny pushes, at right angles to the door, with a force of 132 N at a point 83 cm from the central axis. Show that she exerts a torque of 110 Nm. Show that the rotational inertia of the door is 230 kgm 2. (c) Calculate the rotational kinetic energy gained by the door from Jenny. Explain how this gain in energy is related to the force Jenny exerted on the door. (e) Dorothy tells Jenny that by pushing at B she can get the door rotating to the same speed in the same time with less force. Discuss whether this idea is correct. (f) (g) The door has to rotate through a total angular displacement of 2.0 rad to allow Jenny to walk through the door (above diagram). This total includes the angular displacement during acceleration and the angular displacement at constant angular velocity. Calculate the angular displacement of the door, from the instant it reaches a constant angular velocity, until it has rotated through a total of 2.0 rad. Show that the total time that Jenny is inside the revolving door (from the moment she starts pushing until she exits), can be expressed as an equation: where ω is the maximum angular velocity of the door, α is the angular acceleration of the door and θ the angle through which the door rotates from when she stops pushing until when she exits.

5 SIMPLE HARMONIC MOTION (2008;1) Chairlifts carry skiers to the top of the mountain by way of a continuouslymoving steel cable to which the chairs are attached. As soon as the skiers sit down, the chair is lifted from the ground by the cable. The chair swings back and forth for a while. The diagram shows a side view of a chair as it oscillates (amplitude not to scale). The swinging motion of the chair and skier approximates to a simple pendulum 2.1 m long with a mass of 2.0 x 10 2 kg. Acceleration due to gravity = 9.81 ms -2 (c) The moving cable carries the chair up the hill, and so it has translational motion as well as swinging motion. The swinging motion of the chair is parallel to its translational motion, as shown in the diagram. The pendulum swing of the chair has amplitude 0.69 m. The average translational speed of the chair is 1.2 ms -1. Calculate the maximum speed at which the chair can travel past a stationary observer on the snow. Mike is walking slowly down the hill at 0.20 ms -1. At one instant the chair is moving at exactly the same velocity as Mike. Determine a position of the chair where this occurs. Describe this position by calculating its angle relative to the phasor shown, which represents an equilibrium position. This swinging is uncomfortable, so the chairs are designed to stop swinging within about four oscillations. Sketch a graph to show how the displacement of the chair (relative to the pendulum motion equilibrium position) varies with time. Show FOUR complete oscillations from the moment the chair starts to swing.

6 GOLF IN SPACE (2008;4) In November 2006, flight engineer Mikhail Tyurin hit a golf ball while he was in space, orbiting Earth on a mission on the International Space Station. When Tyurin took the shot from outside of the space craft, his feet were held by his colleague. Describe what sort of motion would have occurred if he had taken the golf shot while he was floating freely in orbit. QUESTION TWO (2007;2) Acceleration due to gravity = 9.81 ms -2. A slowly moving bowl will travel in a curved path, not a straight path. The curved path of the bowl is due to its slightly non-spherical shape, which is created by shaving a small amount off one side of the bowl. The diagrams below show the shape of the bowl. The diagram below shows the bowl, viewed from the front, held stationary on a flat surface. The reaction force, F R, on the bowl acts from the point of contact, P, through the centre of curvature of the spherical part. The gravity force, F, acts through the centre of mass of the bowl. (a) State why the centre of mass is not in the same position as the centre of curvature.

7 If the bowl is to stay in the position shown, it must be held there. If it is released it will roll over sideways, as shown in the diagram below, to position B. Explain why, when it is released, the bowl will not stay balanced at position A. When the bowl reaches position B, it has both rotational and linear kinetic energy. At position B, the bowl has a rotational speed of 2.8 rads -1. (c) The rotational inertia of the bowl is 2.16 x 10-3 kgm 2. Show that the rotational kinetic energy of the bowl at position B is 8.5 mj. The mass of the bowl is 1.50 kg and the radius, r, of its spherical surface is m. Show that the linear kinetic energy of the bowl when it reaches position B is 21 mj (You may assume that the centre of mass travels in an approximately horizontal line). In fact, the centre of mass of the bowl drops slightly as shown in the diagram. (e) (f) At position B, where has the linear and rotational energy come from? Calculate the distance the centre of mass of the bowl drops as it rolls from position A to position B (Ignore any energy losses). While the bowl is rolling over sideways, friction between the bowl and the flat surface is acting in the direction shown below. (g) (h) Explain why friction acts in this direction. When the bowl is rolling forwards (after it has been bowled), this frictional force will still be acting sideways (at right angles to the forward direction). Explain why, after it has been bowled, the path of the bowl curves.

8 DIVING OFF THE HIGH BOARD (2006;3) Hopi performs a dive from the high board. After B leaving the board at A, he travels up in the air to B, tucking his body into a ball. At the end of his dive, he straightens his body and enters the water head first at C. When Hopi s body is in the tucked position during the rotations, his rotational inertia is 3.73 kgm 2. Hopi s mass is 76 kg. (a) When Hopi s body is in the tucked position, his shape can be modelled by a solid sphere of rotational inertia I = 2 5 mr2 Calculate the radius of the sphere that models Hopi s shape. Explain why Hopi must tuck his body if the rotations are to be completed before he enters the water. While his body is in the tucked position, Hopi s angular speed is a constant 9.82 rads -1. He does two complete rotations in this tucked position. Show that his angular momentum, while he is rotating in the tucked position, is 36.6 kgm 2 s -1. (c) Calculate the time it takes him to complete the two rotations in the tucked position. At the end of the tucked rotations, Hopi straightens his body for the entry into the water. What physics principle applies while Hopi is straightening his body? It takes Hopi 0.32 s to straighten his body, and during this time his rotational inertia increases by factor of 5. (e) Calculate his angular deceleration during this time.

9 LINEAR AND ROTATIONAL MOTION (2005;1) The London Eye is a giant rotating wheel that has 32 capsules attached at evenly spaced intervals to its outer rim. Passengers riding in the capsules get spectacular views over London, especially at the top. The capsules each have mass 1.0 x 10 4 kg and are at a distance of 68 m from the centre of the wheel. They travel at a constant speed of 0.26 ms -1. (a) Show that the angular speed of the wheel is 3.8 x 10-3 rads Calculate the time it takes the wheel to travel a complete revolution. When the London Eye is started up each day there are no passengers in the capsules and it takes2.3 s for an average net torque of 4.6 x 10 7 Nm to accelerate the wheel from rest to its operational speed of 0.26 ms -1. (c) (e) (f) (g) (h) Calculate the average angular acceleration of the wheel. Calculate the angle, in degrees, the wheel turns through during start-up. Calculate the rotational inertia of the wheel. In practice, the angular acceleration will gradually reduce to zero during the 2.3 s. It will continue to be zero after this time even though a torque continues to be applied. Explain why the angular acceleration behaves in this way. At what point on the wheel could a force be applied to give maximum torque? By considering what happens to kinetic energy, gravitational potential energy and any other forms of energy, describe and explain the energy transformations that occur while the wheel rotates. The passengers enter and leave the capsules from a deck at the bottom of the wheel. Because the wheel does not stop, each capsule is travelling in the arc of a circle as it moves past the horizontal deck. It takes 29 s for a capsule to pass in front of the deck. The first part of the deck, which is passed during the first half of this time, is for passengers leaving the capsule. The second part of the deck is used by passengers entering the capsule.

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