Concept explainers
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on a horizontal section of a track as shown in Figure P10.33. It rolls around the inside of a vertical circular loop of radius r = 45.0 cm. As the ball nears the bottom of the loop, the shape of the track deviates from a perfect circle so that the ball leaves the track at a point h = 20.0 cm below the horizontal section. (a) Find the ball’s speed at the top of the loop. (b) Demonstrate that the ball will not fall from the track at the top of the loop. (c) Find the ball’s speed as it leaves the track at the bottom. (d) What If? Suppose that static friction between ball and track were negligible so that the ball slid instead of rolling. Describe the speed of the ball at the top of the loop in this situation. (e) Explain your answer to part (d).
Figure P10.33
(a)
The speed of the ball at the top of the loop.
Answer to Problem 33P
The speed of the ball at the top of the loop is
Explanation of Solution
The radius of the circular loop is
Write the expression for the law of energy conservation between horizontal track and the top of the loop
Here,
Write the expression for the translational kinetic energy of the ball at horizontal track
Here,
Write the expression for the rotational kinetic energy of the ball at horizontal track
Here,
Write the expression for the moment of inertia of the ball
Write the expression for the initial angular speed of the ball
Substitute
Write the expression for the translational kinetic energy of the ball at top point of the loop
Here,
Write the expression for the rotational kinetic energy of the ball at top point of the loop
Here,
Write the expression for the moment of inertia of the ball
Write the expression for the initial angular speed of the ball
Substitute
Write the expression for the potential energy of the ball at top point of the loop
Here,
Substitute
Simplify the above equation for
Conclusion:
Substitute
Therefore, the speed of the ball at the top of the loop is
(b)
The reason that the ball will not fall from the track at the top of the loop.
Answer to Problem 33P
The ball will not fall because the value of the centripetal acceleration is more than the acceleration due to gravity at the top point of the circular loop.
Explanation of Solution
The radius of the circular loop is
Formula to calculate the centripetal acceleration on the ball at the top of the loop
Here,
Substitute
Thus, the centripetal acceleration act on the ball at the top of the loop is
Since the centripetal acceleration at the top of the loop is more than the acceleration due to gravity that is
Conclusion:
Therefore, the ball will not fall because the value of the centripetal acceleration is more than the acceleration due to gravity at the top point of the circular loop.
(c)
The speed of the ball as it leaves the track at the bottom.
Answer to Problem 33P
The speed of the ball as it leaves the track at the bottom is
Explanation of Solution
The radius of the circular loop is
Write the expression for the law of energy conservation between horizontal track and the bottom of the loop
Here,
Write the expression for the translational kinetic energy of the ball at bottom point of the loop
Here,
Write the expression for the rotational kinetic energy of the ball at bottom point of the loop
Here,
Write the expression for the initial angular speed of the ball
Substitute
Write the expression for the potential energy of the ball at top point of the loop
Here,
Substitute
Simplify the above equation for
Conclusion:
Substitute
Thus, the speed of the ball as it leaves the track at the bottom is
(d)
The speed of the ball at the top of the loop if ball slide instead of roll.
Answer to Problem 33P
The speed of the ball at the top of the loop is imaginary.
Explanation of Solution
Write the expression for the law of energy conservation between horizontal track and the top of the loop
Here,
Write the expression for the rotational kinetic energy of the ball at horizontal track
Write the expression for the new translational kinetic energy of the ball at top point of the loop
Here,
Write the expression for the new potential energy of the ball at top point of the loop
Substitute
Simplify the above equation for
Substitute
Since the value inside the square root is negative that means the value is imaginary. This condition is impractical.
Thus, the speed of the ball at the top of the loop is imaginary.
Conclusion:
Therefore, the speed of the ball at the top of the loop is imaginary. It can’t be calculated.
(e)
The explanation of the solution of part (d).
Answer to Problem 33P
The ball has not sufficient energy to arrive at top of the circular loop.
Explanation of Solution
The velocity comes out to be imaginary in part (d) that indicates the situation in impossible because as the boll slide instead of rolling, the ball have only translational kinetic energy which is insufficient for the ball to reach the top point on the circular loop.
Thus, the ball did not arrive at the top point of the loop.
Conclusion:
Therefore, the ball has not sufficient energy to arrive at top of the circular loop.
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Chapter 10 Solutions
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- A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on a horizontal section of a track as shown in Figure P10.33. It rolls around the inside of a vertical circular loop of radius r = 45.0 cm. As the ball nears the bottom of the loop, the shape of the track deviates from a perfect circle so that the ball leaves the track at a point h = 20.0 cm below the horizontal section. (a) Find the balls speed at the top of the loop. (b) Demonstrate that the ball will not fall from the track at the top of the loop. (c) Find the balls speed as it leaves the track at the bottom. (d) What If? Suppose that static friction between ball and track were negligible so that the ball slid instead of rolling. Describe the speed of the ball at the top of the loop in this situation. (e) Explain your answer to part (d). Figure P10.33arrow_forwardA solid sphere of mass m and radius r rolls without slipping along the track shown in Figure P10.83. It starts from rest with the lowest point of the sphere at height h above the bottom of the loop of radius R, much larger than r. (a) What is the minimum value of h (in terms of R) such that the sphere completes the loop? (b) What are the force components on the sphere at the point P if h = 3R? Figure P10.83arrow_forwardAn electric motor turns a flywheel through a drive belt that joins a pulley on the motor and a pulley that is rigidly attached to the flywheel as shown in Figure P10.37. The flywheel is a solid disk with a mass of 80.0 kg and a radius R = 0.625 m. It turns on a frictionless axle. Its pulley has much smaller mass and a radius of r = 0.230 m. The tension Tu in the upper (taut) segment of the belt is 135 N, and the flywheel has a clockwise angular acceleration of 1.67 rad/s2. Find the tension in the lower (slack) segment of the belt. Figure P10.37arrow_forward
- Review. An object with a mass of m = 5.10 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.250 m and mass M = 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center as shown in Figure P10.45. The suspended object is released from rest 6.00 m above the floor. Determine (a) the tension in the string, (b) the acceleration of the object, and (c) the speed with which the object hits the floor. (d) Verify your answer to part (c) by using the isolated system (energy) model. Figure P10.45arrow_forwardThe reel shown in Figure P10.71 has radius R and moment of inertia I. One end of the block of mass m is connected to a spring of force constant k, and the other end is fastened to a cord wrapped around the reel. The reel axle and the incline are frictionless. The reel is wound counterclockwise so that the spring stretches a distance d from its unstretched position and the reel is then released from rest. Find the angular speed of the reel when the spring is again unstretched. Figure P10.71arrow_forwardA disk with moment of inertia I1 rotates about a frictionless, vertical axle with angular speed i. A second disk, this one having moment of inertia I2 and initially not rotating, drops onto the first disk (Fig. P10.50). Because of friction between the surfaces, the two eventually reach the same angular speed f. (a) Calculate f. (b) Calculate the ratio of the final to the initial rotational energy. Figure P10.50arrow_forward
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