A horizontal 873 N merry-go-round of radius 1.11 m is started from rest by a constant horizontal force of 70.6 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 3.59 s. The acceleration of gravity is 9.8 m/s2 . Assume the merry-go round is a solid cylinder.

Answer in units of J.

is 721.12 ... use KE=(((F*t)^2)g)/W W= 873, F=70.6 t=3.59, g=9.8

math

To find the kinetic energy of the merry-go-round, we can use the equation:

Kinetic energy (KE) = 1/2 * I * ω^2

where I is the moment of inertia and ω is the angular velocity.

First, we need to find the angular acceleration (α). The torque (τ) acting on the merry-go-round can be calculated using the equation:

τ = r * F

where r is the radius and F is the force applied tangentially.

Given:
r = 1.11 m
F = 70.6 N

τ = r * F
= 1.11 m * 70.6 N
= 78.366 N*m

The equation for torque is:
τ = I * α

Rearranging the equation, we get:
α = τ / I

We can find I using the moment of inertia equation for a solid cylinder:

I = (1/2) * m * r^2

where m is the mass of the merry-go-round.

Given:
m = F / g
= 70.6 N / 9.8 m/s^2
= 7.204 kg (approx.)

I = (1/2) * 7.204 kg * (1.11 m)^2
= 4.118 kg*m^2 (approx.)

Now, we can find α:
α = 78.366 N*m / 4.118 kg*m^2
= 19.0 rad/s^2 (approx.)

Next, we need to find the angular velocity (ω) after 3.59 s. We can use the equation:

ω = ω0 + α * t

where ω0 is the initial angular velocity, t is the time.

Given:
ω0 = 0 rad/s (starting from rest)
t = 3.59 s

ω = 0 rad/s + (19.0 rad/s^2) * (3.59 s)
= 68.41 rad/s (approx.)

Now, we can calculate the kinetic energy using the formula:

KE = 1/2 * I * ω^2

KE = (1/2) * 4.118 kg*m^2 * (68.41 rad/s)^2
= 9070 J (approx.)

Therefore, the kinetic energy of the merry-go-round after 3.59 s is approximately 9070 J.

To find the kinetic energy of the merry-go-round, we first need to determine its angular acceleration and angular velocity.

The torque applied to the merry-go-round is equal to the product of its moment of inertia and angular acceleration, given by the equation:

τ = I α

where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

For a solid cylinder rotating about its central axis, the moment of inertia is given by the equation:

I = 1/2 m r^2

where m is the mass of the merry-go-round and r is its radius.

Given that the radius of the merry-go-round is 1.11 m and the mass is not provided, we cannot directly find the moment of inertia. However, we can solve for it using the known force applied tangentially to the merry-go-round.

The torque applied to the merry-go-round is also given by the equation:

τ = r F

where F is the force applied tangentially to the merry-go-round and r is the radius.

Substituting the given force of 70.6 N and radius of 1.11 m into the equation, we can find the torque:

τ = (1.11 m) (70.6 N) = 78.126 N·m

Now we can equate the torque to the product of the moment of inertia and angular acceleration:

78.126 N·m = (1/2 m r^2) α

Simplifying the equation, we have:

α = (2 τ) / (m r^2)

Given that the torque is 78.126 N·m and the radius is 1.11 m, we need the mass of the merry-go-round to find the angular acceleration.

Unfortunately, the mass is not given in the problem statement. If you have the mass value, you can proceed with the calculations.