A 190-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.700 rev/s in 2.00 s? (State the magnitude of the force.)
wrong
To find the constant force required to bring the merry-go-round from rest to an angular speed of 0.700 rev/s, we can use the equation:
τ = I * α
Where
τ is the torque (force applied * radius),
I is the moment of inertia of the disk,
and α is the angular acceleration.
The moment of inertia for a solid disk rotating about an axis through its center is given by:
I = (1/2) * m * r^2
Where
m is the mass of the disk,
and r is the radius of the disk.
First, let's calculate the moment of inertia:
I = (1/2) * m * r^2
= (1/2) * 190 kg * (1.50 m)^2
= 213.75 kg·m²
Next, let's calculate the angular acceleration:
α = ω / t
Where
ω is the change in angular speed,
and t is the time taken.
ω = (0.700 rev/s) - 0 rev/s
= 0.700 rev/s
t = 2.00 s
α = (0.700 rev/s) / 2.00 s
= 0.350 rev/(s²)
To find the torque:
τ = I * α
= 213.75 kg·m² * (0.350 rev/(s²))
= 74.81 kg·m²/s²
Lastly, let's determine the force required:
τ = F * r
Where
F is the force applied,
and r is the radius of the disk.
F = τ / r
= (74.81 kg·m²/s²) / (1.50 m)
≈ 49.87 N
Therefore, the constant force required to bring the merry-go-round from rest to an angular speed of 0.700 rev/s in 2.00 s is approximately 49.87 N.
To find the constant force required to bring the merry-go-round to the desired angular speed, we can use the principle of angular acceleration.
The formula for angular acceleration is:
α = (ωf - ωi) / t,
where:
α is the angular acceleration,
ωf is the final angular speed,
ωi is the initial angular speed, and
t is the time taken to reach the final angular speed.
Given:
The initial angular speed ωi is 0, as the merry-go-round is at rest.
The final angular speed ωf is 0.700 rev/s.
The time taken t is 2.00 s.
Substituting these values into the formula, we have:
α = (0.700 rev/s - 0 rev/s) / 2.00 s.
Now, we need to convert the final angular speed from revolutions per second to radians per second, as the formula requires angular speed to be in radians. Since 1 revolution equals 2π radians, we can convert the final angular speed as follows:
ωf = 0.700 rev/s * 2π rad/rev.
Now we can calculate the angular acceleration:
α = (0.700 rev/s * 2π rad/rev - 0 rev/s) / 2.00 s.
Simplifying:
α = (1.40π rad/s) / 2.00 s.
α = 0.70π rad/s².
Next, we can use the formula for torque:
τ = I * α,
where:
τ is the torque,
I is the moment of inertia, and
α is the angular acceleration.
The moment of inertia of a uniform, solid disk is given by the formula:
I = (1/2) * M * R^2,
where:
M is the mass of the disk, and
R is the radius of the disk.
Given:
The mass M is 190 kg.
The radius R is 1.50 m.
Substituting these values into the moment of inertia formula, we have:
I = (1/2) * 190 kg * (1.50 m)^2.
Simplifying:
I = 1/2 * 190 kg * 2.25 m^2.
I = 213.75 kg·m².
Finally, we can find the torque:
τ = 213.75 kg·m² * 0.70π rad/s².
τ ≈ 470.25π N·m.
Since torque is equal to the force multiplied by the radius, we can write:
τ = F * R,
where:
F is the force, and
R is the radius.
Solving for the force:
F = τ / R.
Substituting the values, we have:
F = (470.25π N·m) / 1.50 m.
F ≈ 314π N.
The magnitude of the force required to bring the merry-go-round to the desired angular speed is approximately 314π N.
F*r = I * a
Where F is the force on the merry-go-round, r is the radius, I is the moment of inertia of the merry-go-round, and a is the angular acceleration
for a solid, horizontal disk, the moment of inertia is: 1/2*m*r^2
F*r = 1/2*m*r^2*a
F = 1/2*m*r*a
w = w0 + a*t
where w is the angular speed, t is time, w0 is the initial angular speed
0.7 rev / s *(2*pi radians/rev) = 4.3 radians/s
4.3 = a*2
a = 2.2 rad/s^2
Plug this into the above equation to solve for F