Estimate the force a person must exert on a string attached to a 0.160 kg ball to make the ball revolve in a circle when the length of the string is 0.600 m. The ball makes 1.80 revolutions per second. Do not ignore the weight of the ball. In particular, find the magnitude of FT, and the angle it makes with the horizontal. [Hint: Set the horizontal component of FT equal to maR; also, since there is no vertical motion, what can you say about the vertical component of FT?] ***I'VE FOUND THE FORCE; I JUST NEED HELP FINDING THE ANGLE***
The vertical component of the tension equals the weight
The angle of the string relative to horizontal is arctan(W/Ft)
To find the angle that the force FT makes with the horizontal, we can use trigonometry.
Let's consider the forces acting on the ball in the vertical direction. Since there is no vertical motion, the vertical component of FT must balance the weight of the ball.
The weight of the ball can be calculated as W = mass * gravity, where gravity is approximately 9.8 m/s^2.
W = (0.160 kg) * (9.8 m/s^2) = 1.568 N
Since the vertical component of FT balances the weight, we have:
FT_vertical = -1.568 N (negative because it is opposite in direction to the weight)
Now, let's consider the forces acting on the ball in the horizontal direction. The centripetal force is responsible for the circular motion of the ball, and it can be calculated using the formula:
centripetal force = mass * (velocity^2 / radius)
The velocity of the ball can be found by multiplying the circumference of the circle by the number of revolutions per second and the radius:
velocity = 2π * radius * (1.80 rev/s) = 2π * 0.600 m * (1.80 rev/s)
The circumference of the circle is given by 2π * radius.
Now, let's calculate the centripetal force:
centripetal force = (0.160 kg) * ((2π * 0.600 m * (1.80 rev/s))^2 / 0.600 m) = 5.758 N
The horizontal component of FT must equal the centripetal force:
FT_horizontal = 5.758 N
Now, we can find the angle θ that FT makes with the horizontal using trigonometry:
tan(θ) = FT_vertical / FT_horizontal
θ = arctan(FT_vertical / FT_horizontal)
θ = arctan(-1.568 N / 5.758 N)
θ ≈ -15.89°
Therefore, the angle that FT makes with the horizontal is approximately -15.89°. Note that the negative sign indicates that the force is directed below the horizontal.
To find the angle made by the force applied, you can use trigonometry. Let's break down the problem and use the hint provided.
Given:
- Mass of the ball (m) = 0.160 kg
- Length of the string (r) = 0.600 m
- Revolutions per second (ω) = 1.80 rev/s
We need to find the angle (θ) that the force makes with the horizontal.
1. First, find the angular velocity (ω) in radians/second. Since there are 2π radians in one revolution, multiply the number of revolutions per second by 2π to convert it into radians/second:
ω = 1.80 rev/s * 2π rad/rev
2. Calculate the velocity of the ball:
The velocity of the ball (v) when moving in a circular path is given by: v = ω * r
Plug in the values:
v = (1.80 rev/s * 2π rad/rev) * 0.600 m
3. Determine the net force (Fnet) required to maintain circular motion:
The net force is the centripetal force (Fnet = m * a_c), where a_c is the centripetal acceleration.
The centripetal acceleration is given by: a_c = v^2 / r
Substitute the values to find Fnet:
Fnet = m * (v^2 / r)
4. Decompose the net force (Fnet) into its components:
The horizontal component of the force (FT) can be calculated as:
FT = Fnet * cos(θ), where θ is the angle the force makes with the horizontal.
5. Substitute the value of Fnet from step 3 into the equation for FT:
FT = m * (v^2 / r) * cos(θ)
Now, we have the equation for FT. To find the angle θ, we can rearrange the equation.
6. Multiply both sides of the equation by r and divide by m * v^2:
r * FT / (m * v^2) = cos(θ)
7. Take the inverse cosine (arccos) of both sides to solve for θ:
θ = arccos(r * FT / (m * v^2))
Now, you can plug in the known values into this equation to find the angle θ.