Homework 8 Problems: Rotational Motion 1. A machine part rotates at an angular speed of 0.06 rad/s; its angular speed is then increased to 2.2 rad/s at an angular acceleration of 0.70 rad/s2. a. [5 points] Find the angle through which the part rotates before reaching this final speed. b. [5 points] In general, if both the initial and final angular speed are doubled at the same angular acceleration, by what factor is the angular displacement changed? Why? Answer: (a) To determine the final angular position, we use rotational kinematics equation 3 ( ) ( ( ) ) (b) Based on rotational kinematics equation 3, we see that is proportional to . Therefore, doubling the final and initial speed (while keeping the angular acceleration constant) will quadruple the angular displacement. 2. A car initially traveling at 29.0 m/s undergoes a constant negative acceleration of magnitude 1.75 m/s2 after its brakes are applied. a. [7 points] How many revolutions does each tire make before the car comes to a stop, assuming the car does not skid and the tires have radii of 0.330 m? b. [3 points] What is the angular speed of the wheels when the car has traveled half the total distance? Answer: (a) To determine the amount of revolutions each tire makes, we need to convert linear quantities to angular quantities. The final angular speed is given by The angular acceleration is given by To determine the number of revolutions made, we use kinematics equation 2 ( ) ( ) (b) If the car travels half the total distance, this implies that the wheels will rotate half the total revolutions. Assuming the same deceleration, the angular speed of the wheels when the car has traveled half the total distance is given by ( √ ( ) √( ) ) ( )( ) 3. A pail of water is rotated in a vertical circle of radius 1.00 m. a. [3 points] What two external forces act on the water in the pail? Which of the two forces is most important in causing the water to move in a circle? b. [4 points] What is the pail’s minimum speed at the top of the circle if no water is to spill out? c. [3 points] If the pail with the speed found in part (b) were to suddenly disappear at the top of the circle, describe the subsequent motion of the water. Would it differ from the motion of a projectile? Answer: (a) For a pail of water moving in a circle, there are two forces that operate on the water: the weight of the water itself (i.e. the gravitational force on the water) and the normal force from the pail acting on the water. The normal force of the bottom of the pail is most important in causing the water to move in a circle since the normal force always points towards the center of the circle for a pail of water moving in a vertical circle. (b) At the top of the circle, the gravitational force pulls the water toward the center of the circular path. If the water is just on the verge of spilling out, this means that the normal force between the pail and the water is zero. Thus, the only force on the water must be the gravitational force Solving for the minimum speed gives √ √( )( ) (c) If the pail was removed, then there is no longer a centripetal force that causes the water to move in a circular path. Therefore, the water will initially move in the direction of the tangential velocity at that point. Since the water is in the air, the water will travel like a projectile with a horizontal initial velocity of 3.13 m/s. 4. Because of Earth’s rotation about its axis, a point on the equator has a centripetal acceleration of 0.0340 m/s2, whereas a point at the poles has no centripetal acceleration. a. [7 points] Show that, at the equator, the gravitational force on an object (i.e. the object’s true weight) must exceed the object’s apparent weight. b. [3 points] What are the apparent weights of a 75.0-kg person at the equator and at the poles? (Assume Earth is a uniform sphere and take g = 9.800 m/s2.) Answer: (a) The apparent weight of an object (as measured on a scale) is the magnitude of the normal force. Taking the earth as uniform sphere, Newton’s 2nd law in the radial direction gives ∑ Here, ω is the angular rotation speed of the earth. When an object is at the poles, the distance from the object to the axis of rotation is zero and thus there is no centripetal acceleration. This implies that the object’s apparent weight is equal to the gravitational force on the object (i.e. the object’s true weight). However, when an object is at the equator, the distance from the object to the axis of rotation is no longer zero and thus there is a centripetal acceleration. This indicates that there is an imbalance of the forces in the radial direction, giving an apparent weight of ( This implies that the object’s apparent weight. ) and thus, the gravitational force on the object exceeds This can also be explained via the centrifugal force. For a person at the pole, the centrifugal force is zero and thus, the apparent weight is the same as the person’s true weight. For a person at the equator, the centrifugal force points outward (i.e. away from Earth’s surface) and this will counterbalance the person’s true weight. Consequently, the apparent weight will be less than the true weight of the person. (b) For a 75.0-kg person, the apparent weight of the man at the poles is ( )( ) whereas the apparent weight of the man at the equator is ( ) ( )( ) 5. One end of a cord is fixed and a small 0.500-kg object is attached to the other end, where it swings in a section of a vertical circle of radius 2.00 m, as shown below. When θ = 20.0°, the speed of the object is 8.00 m/s. a. [4 points] Find the tension in the string, b. [3 points] Find the tangential and radial components of acceleration. Find the total acceleration. c. [3 points] Is your answer changed if the object is swinging down toward its lowest point instead of swinging up? Explain your answer. Answer: (a) Here the tension force serves as the centripetal force necessary for this object to move in a circular path. Using a coordinate system in which the x-axis points in the tangential direction and the y-axis points in the radial direction, we can apply Newton’s 2nd law to determine the tension. Using Newton’s 2nd law in the y-direction, we have Solving for the tension force ( ) ( gives ( ) )( ( ) ( ) ) (b) The centripetal acceleration can be calculated from the object’s speed when θ = 20.0°: ( ) The tangential acceleration can be calculated from Newton’s 2nd law in the x-direction Solving for the tangential acceleration gives ( ) The total acceleration is given by √ √( ) ( ) (c) A schematic diagram of this problem is given below If the object is swinging down instead of up, this would imply that the tangential acceleration would be positive. However, the magnitude of the tangential acceleration would be the same since the angle θ is the same. This would also imply that the speed is the same, which further implies that the centripetal acceleration is the same. Thus, the total acceleration would remain the same. Bonus: [7 points] A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane with the cord making a 30⁰ angle with the vertical. a. Determine the ball’s speed. b. If, instead the ball is revolved so that its speed is 4.0 m/s, what angle does the cord make with the vertical? c. If the cord can withstand a maximum tension of 9.8 N, what is the highest speed at which the ball can move? Answer: (a) The ball is subject to two forces: the gravitational force and the tension force. Here, the horizontal component of the tension force produces a centripetal acceleration, which acts towards the center of the circle. Using Newton’s 2nd law in the radial and vertical direction we have ∑ ∑ Eliminating T in both equations gives If the height from the hand to the plane of the ball is h, we can also write Setting both equations equal to each other gives √ √ Note that the radius of the horizontal circle can be written as from the hand to the plane of the ball h can be written as √ √ √ ( )( ) ( ( and the height . This gives us ) ) (b) If the ball now has a tangential velocity of 4.0 m/s, then we can write the tangential velocity as a function of the radius of the circle ( ) Solving this quadratic equation gives . (c) If the cord can withstand a maximum tension of 9.8 N, then the angle of the cord with the vertical is given by ( ( ( ) )( ) Therefore, the maximum velocity is given as √ √ √ ( )( ) ( ( ) ) )

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