A horizontal 810-N merry-go-round of radius 1.70 m is started from rest by a constant horizontal force of 55 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 2.0 s. (Assume it is a solid cylinder.)
The kinetic energy is (1/2) I w^2 , where I is the moment of inertia, which is (1/2) M R^2 for a solid cylinder. Therefore KE = (1/2) M R^2 w^2
w is the angular velocity in radians per second. To determine its value after 2.0 seconds, you will need to use the equation of motion
T = I*dw/dt
where T is the torque, which in this case is 55 * 1.70 = 93.5 N m/s. You will need to compute dw/dt, the rate of change of angular velocity.
The value of w at 2.0 s is
w (@t=2) = dw/dt * 2.0 radians per second.
Now put all that information together and do the calculation.
To find the kinetic energy of the merry-go-round after 2.0 seconds, we need to first calculate the angular acceleration of the merry-go-round.
The torque applied to the merry-go-round can be found using the equation:
τ = Fr
where τ is the torque, F is the applied force, and r is the radius of the merry-go-round.
Substituting the given values, we have:
τ = (55 N)(1.70 m)
Next, we can use the equation for torque, τ, and the moment of inertia, I, of a solid cylinder to calculate the angular acceleration, α:
τ = Iα
For a solid cylinder, the moment of inertia is given by:
I = (1/2)mr^2
where m is the mass of the merry-go-round.
Substituting the values, we have:
(55 N)(1.70 m) = (1/2)(m)(1.70 m^2)α
Now, let's calculate α:
α = (2(55 N)(1.70 m)) / (m)(1.70 m^2)
Given that m is the mass of the merry-go-round, we don't have a specific value for it in the problem. However, we can use the definition of mass-density, ρ, to relate mass to the volume and density of the material. For a solid cylinder, the mass can be expressed as:
m = ρV
where V is the volume of the cylinder.
Substituting this expression for mass into our equation for α, we have:
α = (2(55 N)(1.70 m)) / ((ρV)(1.70 m^2))
The volume of a cylinder is given by:
V = πr^2h
In this problem, the height, h, is not given, but it is not needed to solve for α. We can eliminate it as it will cancel out in the calculations.
α = (2(55 N)(1.70 m)) / ((ρπr^2)(1.70 m^2))
Simplifying,
α = (2(55 N)(1.70 m)) / ((ρπ)(1.70 m^2))
Now, we can calculate the angular acceleration, α.
Let's assume a value for the density, ρ. For simplicity, let's assume it is the density of steel, which is approximately 7,860 kg/m^3.
Substituting the values for α and the given time, we can find the angular velocity, ω:
ω = αt
Substituting t = 2.0 s, we have:
ω = α × 2.0 s
Finally, we can calculate the kinetic energy, KE, of the merry-go-round using the equation:
KE = (1/2)Iω^2
Substituting the moment of inertia, I, and the calculated angular velocity, ω, we have:
KE = (1/2)((1/2)mr^2)(ω^2)
Substituting m and ω as found in the previous steps, we can now calculate the kinetic energy of the merry-go-round after 2.0 seconds.