Question

A 2 kg ball swings in a vertical circle on the end of an 60-cm-long string. The tension in the string is 20 N when its angle from the highest point on the circle is θ=30∘.

Part A

What is the ball's speed when θ=30∘?

Part B

What is the magnitude of the ball's acceleration when θ=30∘?

Express your answer to two significant figures and include the appropriate units.

Part C

What is the direction of the ball's acceleration when θ=30∘? Give the direction as an angle from the r-axis.

Express your answer to two significant figures and include the appropriate units.

Answer 1

The ball's speed is approximately 4.24 m/s when θ = 30∘. The magnitude of the ball's acceleration is approximately 31.6 m/s^2 when θ = 30∘. The direction of the ball's acceleration is 180∘ when θ = 30∘.

Step-by-step explanation:

Part A: To find the ball's speed when θ = 30∘, we can use the concept of conservation of energy. At the highest point of the circle, all of the ball's gravitational potential energy is converted into kinetic energy, given by the equation:

mgh = 1/2 mv^2

Where m is the mass of the ball, g is the acceleration due to gravity, h is the height of the highest point (which can be calculated using the length of the string and the angle θ), and v is the speed of the ball. Rearranging the equation, we can solve for v:

v = √(2gh)

Substituting the given values, we have:

v = √(2 * 9.8 m/s^2 * 0.6 m * (1 - cos(30∘)))

Calculating this expression, we find that the ball's speed when θ = 30∘ is approximately 4.24 m/s.

Part B: The magnitude of the ball's acceleration can be determined using the centripetal acceleration formula:

a = v^2/r

Where a is the acceleration, v is the speed of the ball, and r is the radius of the circle (equal to the length of the string).

Substituting the given values, we have:

a = (4.24 m/s)^2 / 0.6 m

Calculating this expression, we find that the magnitude of the ball's acceleration when θ = 30∘ is approximately 31.6 m/s^2.

Part C: The direction of the ball's acceleration when θ = 30∘ can be determined from the direction of the net force acting on the ball. This force is directed towards the center of the circle, which corresponds to an angle of 180∘ from the positive x-axis. Therefore, the direction of the ball's acceleration is 180∘.

Answer 2

The ball's speed when θ = 30∘ is approximately 1.73 m/s. The magnitude of the ball's acceleration when θ = 30∘ is approximately 5 m/s². The direction of the ball's acceleration when θ = 30∘ is 0∘ from the r-axis.

Step-by-step explanation:

When the ball is at an angle of 30∘, the tension in the string is equal to the centripetal force acting on the ball.

Using the equation for centripetal force:

Fc = mv² / r

where Fc is the centripetal force, m is the mass of the ball, v is the velocity of the ball, and r is the radius of the circle.

In this case, the radius of the circle is the length of the string, which is 60 cm (or 0.6 m).

Substituting the given values into the equation:

20 N = (2 kg) v² / 0.6 m

Solving for v:

v² = (20 N) * (0.6 m) / (2 kg)

v² = 6 N * m / kg

v = sqrt(6 N * m / kg)

v ≈ 1.73 m/s

So, the ball's speed when θ = 30∘ is approximately 1.73 m/s.

The magnitude of the ball's acceleration when θ = 30∘ can be found using the equation for centripetal acceleration:

ac = v² / r

where ac is the centripetal acceleration.

Substituting the known values:

ac = (1.73 m/s)² / 0.6 m

ac ≈ 5 m/s²

So, the magnitude of the ball's acceleration when θ = 30∘ is approximately 5 m/s².

Part C:

The direction of the ball's acceleration when θ = 30∘ is towards the center of the circle.

The direction can be represented as an angle from the r-axis, which is the vertical axis in this case.

Since the acceleration is directed towards the center of the circle, the angle is 0∘.

So, the direction of the ball's acceleration when θ = 30∘ is 0∘ from the r-axis.