How do I answer these physics questions and what answer should I get?

Question

A 0.5 kg ball that is tied to the end of a 1.5 m light string is revolved in a horizontal plane, with

the string making a 30 o angle with the vertical.

(a) Draw the free-body diagram for a conical pendulum.

(b) Set the equations of motion for the ball.

(c) What is the ball’s speed?

(d) If the ball is revolved with 4.0 m/s speed, what angle the string make with the vertical?

(e) It the string can withstand a maximum tension of 9.8 N, what is the highest speed at which

the ball can revolve?

Answer 1

Let

m = 0.5 kg be the mass of the ball,

L = 1.5 m be the length of the string,

θ = 30° is the angle the string makes with the vertical,

g = 9.81 m/s² be gravitational acceleration,

r (unknown) be the radius of the circle,

v (unknown) be the velocity of the ball.

(a)

There are three forces acting on the ball:

G: downward force of gravity,

T: upward force of tension along the string,

F: horizontal centripetal force toward the center of rotation.

It is important in the free-body diagram that

F = G + T,

which is vector addition of the forces.

It says that G and T are responsible for the centripetal force needed for the circular motion.

(b)

F = mv²/r (1)

(c)

Use equation (1).

To calculate r:

r = Lsin(θ)

To calculate F:

From the free-body diagram in (a) showing F = G + T,

F = Gtan(θ) = mgtan(θ)

Substituting into (1):

mgtan(θ) = mv²/(Lsin(θ)) (2)

v = √(gLtan(θ)sin(θ)) (3)

Substituting actual numbers

v = √(9.81 × 1.5 × tan(30°) × sin(30°)) = 2.061 m/s

(d)

New given: v = 4 m/s

New unknown: θ

Rearranging equation (2)

tan(θ)sin(θ) = v²/gL

Let's denote

U = v²/gL = 4²/(9.81×1.5) = 1.08732585

Then we need to solve

tan(θ)sin(θ) = U

Rewriting in terms of cos(θ) and then solving a quadratic equation over cos(θ):

(sin(θ)/cos(θ)) sin(θ) = U

sin²(θ) = Ucos(θ)

1 - cos²(θ) = Ucos(θ)

cos²(θ) + Ucos(θ) - 1 = 0

cos(θ) = (-U ± √(U² + 4))/2

Approximating U = 1

cos(θ) = (-1 ± √(1² + 4))/2 = (-1 ± √5)/2

Only cos(θ) = (-1 + √5)/2 gives a solution within the range [-1,+1] of cos(θ).

cos(θ) = 0.618

θ = 51.8°

(e)

New given: T = 9.8 N

New unknown: v

From the free-body diagram

cos(θ) = G/T

Substituting actual numbers to get θ:

cos(θ) = mg/T = 0.5 × 9.8 / 9.81 = 0.5

θ = 60°

To calculate v use equation (3)

v = √(gLtan(θ)sin(θ))

= √(9.81 × 1.5 × tan(60°) × sin(60°)) = 4.7 m/s