The rope of a swing is 3.2 m long. Calculate the angle from the vertical at which a 95-kg man must begin to swing in order to have the same kinetic energy at the bottom as a 1500-kg car moving at 1.07 m/s (2.39 mi/hr).

Ke car = (1/2) m v^2

potential energy at top of swing = kinetic energy at bottom of swing

m g h = m g L (1-cos theta)

so
(1/2)(1500)(1.07)^2 = 95 * 9.8 * (1 - cos T)
solve for T

30

To solve this problem, we need to consider the conservation of energy. At the top of the swing (highest point), the potential energy of the swing and the man is maximum, while the kinetic energy is zero. At the bottom of the swing (lowest point), the potential energy is zero and the kinetic energy is maximum.

First, we need to find the velocity of the man at the bottom of the swing. We can use the equation for kinetic energy:

K.E. = (1/2) * m * v^2

where K.E. is the kinetic energy, m is the mass, and v is the velocity.

For the car, its mass is 1500 kg and the velocity is given as 1.07 m/s. Plugging these values into the equation, we can solve for the kinetic energy of the car:

K.E. car = (1/2) * 1500 kg * (1.07 m/s)^2

Next, we need to find the velocity at the bottom of the swing for the man. We can equate the kinetic energies of the man and the car:

K.E. man = K.E. car

(1/2) * m * v_man^2 = (1/2) * 1500 kg * (1.07 m/s)^2

Since the mass of the man is given as 95 kg, we can solve for v_man:

(1/2) * 95 kg * v_man^2 = (1/2) * 1500 kg * (1.07 m/s)^2

Now, let's simplify the equation:

95 kg * v_man^2 = 1500 kg * (1.07 m/s)^2

Divide both sides by 95 to isolate v_man:

v_man^2 = (1500 kg * (1.07 m/s)^2) / 95 kg

v_man^2 = 16.7 m^2/s^2

Now take the square root of both sides to find v_man:

v_man = sqrt(16.7 m^2/s^2)

v_man ≈ 4.088 m/s

At the lowest point of the swing, the total energy is completely kinetic. Now, let's calculate the height difference between the lowest point and the highest point of the swing.

The difference in potential energy between the highest and lowest points is equal to the difference in kinetic energy. Therefore:

Potential energy at the highest point = Kinetic energy at the lowest point

m * g * h = (1/2) * m * v_man^2

where m is the mass, g is the acceleration due to gravity, h is the height difference, and v_man is the velocity at the lowest point.

Since we are looking for the angle with the vertical, we can write h in terms of the length of the rope and the angle.

h = L * [1 - cos(θ)]

where L is the length of the rope and θ is the angle from the vertical.

Substituting this equation into the previous equation, we have:

m * g * L * [1 - cos(θ)] = (1/2) * m * v_man^2

Now we can solve for θ:

g * L * [1 - cos(θ)] = (1/2) * v_man^2

Rearranging the equation, we get:

cos(θ) = 1 - (v_man^2) / (2 * g * L)

Finally, we can solve for θ by taking the arccosine (inverse cosine) of both sides of the equation:

θ = arccos(1 - (v_man^2) / (2 * g * L))

Substituting the given values for v_man, g (approximately 9.8 m/s^2), and L (3.2 m), we can now calculate θ:

θ = arccos(1 - (4.088 m/s)^2 / (2 * 9.8 m/s^2 * 3.2 m))

θ ≈ arccos(1 - 0.133).

Evaluating this expression, we find:

θ ≈ arccos(0.867)

θ ≈ 29.5 degrees

Therefore, the man must begin swinging at an angle of approximately 29.5 degrees from the vertical to have the same kinetic energy at the bottom as the car.