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A 0.20-kilogram solid rubber ball is attached to the end of an 0.60-meter length of light thread. The ball is swung in a

horizontal circle. At one point along the circular path, the speed of the ball is 5.0 meters per second. The following diagram
shows a downward view of the horizontal path.
a
0.60 m
5.0 m/s
b
a. Using a dot model, draw and label all the forces exerted on the ball when it is at point b. (3 points)
b. Determine the speed of the ball at point a. Show your work. (4 points)
c. Determine the tension in the thread at point a. Show your work. (4 points)
d. Determine the tension in the thread at point b. Show your work. (4 points)

User Cruces
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1 Answer

2 votes

a. The forces exerted on the ball at point b are: Tension in the thread , Weight of the ball.

b. The speed of the ball at point a is 5.0 m/s.

c. The tension in the thread at point a is 83.3 N.

d. The tension in the thread at point b is 333 N.

a. Using a dot model, draw and label all the forces exerted on the ball when it is at point b.

The dot model of the forces on the ball at point b is shown below:

+----------------+

| Tension in thread |

+----------------+

|

|

v

|

|

+-----+

| Ball |

+-----+

b. Determine the speed of the ball at point a. Show your work.

Since mechanical energy is conserved, the speed of the ball at point a will be the same as its speed at point b, which is 5.0 m/s.

c. Determine the tension in the thread at point a. Show your work.

The tension in the thread at point a is equal to the centripetal force required to keep the ball moving in a circle.

The centripetal force is given by:

F_c = mv^2/r

where:

m is the mass of the ball (20 kg)

v is the speed of the ball (5.0 m/s)

r is the radius of the circle (0.60 m)

Therefore, the tension in the thread at point a is:

F_c = (20 kg)(5.0 m/s)^2 / 0.60 m = 83.3 N

d. Determine the tension in the thread at point b. Show your work.

The tension in the thread at point b is equal to the weight of the ball plus the centripetal force required to keep the ball moving in a circle.

The weight of the ball is given by:

F_g = mg

where:

m is the mass of the ball (20 kg)

g is the acceleration due to gravity (9.81 m/s^2)

Therefore, the tension in the thread at point b is:

F_b = F_g + F_c = (20 kg)(9.81 m/s^2) + (20 kg)(5.0 m/s)^2 / 0.60 m = 3

A 0.20-kilogram solid rubber ball is attached to the end of an 0.60-meter length of-example-1
User Milad Abooali
by
8.1k points