A merry go round in the park has a radius of 1.8 m and a rotational inertia of 900 kg-m^2. A child pushes the merry go round with a constant force of 80 N applied at the edge and parrallel to the edge. A frictional torque of 12 Nm acts at the axle of the merry go round. What is the rotational acceleration of the merry go round if the child stops pushing after 15 seconds? How long will it take for the merry go round to stop turning?
After the child stops pushing, the rotational acceleration is negative. it slows down.
Use the equation
alpha = L/I
where alpha is the angular DEceleration, I is the moment of inertia and L is the frictional torque.
To determine how long it takes to stop. you need to calculate how fast it is going when the pushing force stops. That is a somewhat harder problem, but uses the same equation. You need to calculate the angular acceleration rate while it is being pushed, and multiply that by the time (15 seconds).
Use the same L/I equation, but in this accvelerating case, the torque is the difference of the pushing and frictional torques.
To find the rotational acceleration of the merry-go-round, we can use the following formula:
τnet = Iα
Where:
τnet is the net torque
I is the rotational inertia
α is the rotational acceleration
First, let's find the net torque. The child applies a force of 80 N at the edge of the merry-go-round. The leverage arm is equal to the radius, which is 1.8 m. Therefore, the torque applied by the child is:
τchild = F × d = 80 N × 1.8 m = 144 Nm
The frictional torque acting at the axle is given as 12 Nm. Since the torque due to friction is in the opposite direction to the torque applied by the child, we can write:
τnet = τchild - τfriction
= 144 Nm - 12 Nm
= 132 Nm
Now we can find the rotational acceleration. Rearranging the formula, we have:
α = τnet / I
α = 132 Nm / 900 kg-m^2
≈ 0.147 rad/s^2
Therefore, the rotational acceleration of the merry-go-round is approximately 0.147 rad/s^2.
To find out how long it will take for the merry-go-round to stop turning, we can use the formula:
ω = ω0 + αt
Where:
ω is the final angular velocity (which is zero since the merry-go-round stops turning)
ω0 is the initial angular velocity
α is the rotational acceleration
t is the time taken
Rearranging the formula, we have:
t = (ω - ω0) / α
Since ω is zero (the final angular velocity) and ω0 is the initial angular velocity, we need to find ω0. Using the equation:
ω0 = αt0
Where t0 is the time the child pushes the merry-go-round, we can calculate ω0:
ω0 = αt0
= 0.147 rad/s^2 × 15 s
= 2.205 rad/s
Now, we can calculate the time it takes for the merry-go-round to stop turning:
t = (ω - ω0) / α
= (0 - 2.205 rad/s) / (0.147 rad/s^2)
≈ 15 seconds
Therefore, it will take approximately 15 seconds for the merry-go-round to stop turning after the child stops pushing.
To find the rotational acceleration of the merry go round, we can use the formula:
τ = Iα
where τ is the net torque acting on the merry go round, I is the rotational inertia, and α is the rotational acceleration.
In this case, the net torque acting on the merry go round is the torque produced by the child's push minus the frictional torque:
τ = (80 N)(1.8 m) - 12 Nm
τ = 144 Nm - 12 Nm
τ = 132 Nm
Substituting this into the formula, we have:
132 Nm = (900 kg-m^2)α
Now, let's solve for α:
α = (132 Nm) / (900 kg-m^2)
α ≈ 0.147 rad/s^2
Therefore, the rotational acceleration of the merry go round is approximately 0.147 rad/s^2.
Next, to find out how long it will take for the merry go round to stop turning, we need to calculate the time it takes for the rotational velocity to decrease to zero.
We can use the formula:
ω = ω₀ + αt
where ω is the final angular velocity, ω₀ is the initial angular velocity, α is the rotational acceleration, and t is the time.
Given that the initial angular velocity is zero (as the child stops pushing), the formula simplifies to:
ω = αt
Since we want to find the time it takes for ω to become zero, we set ω to be zero and solve for t:
0 = αt
t = 0 / α
t = 0
Therefore, it will take zero seconds for the merry go round to stop turning since the rotational acceleration is constant throughout the motion.