A constant torque of 22.9 N·m is applied to a grindstone whose moment of inertia is 0.168 kg·m2. Using energy principles, and neglecting friction, find the angular speed after the grindstone has made 17.1 rev, assuming it started
from rest.
Answer in units of rev/s
M=Iε
ε=M/I = 22.9/0.168 = 136.3 rad/s²
2πN=2πn₀t + εt²/2
2πn=2πn₀ + εt
n₀=0 =>
2πN= εt²/2
2πn= εt => t=2πn/ε
N= πn²/ε
n=sqrt {εN/π}=
=sqrt{136.3•17.1/3.14}=27.24 rev/s
Well, let's see. A grindstone getting a constant torque... sounds like a tough job! But don't worry, I'm here to help.
To find the angular speed, we can use energy principles. The formula to work with here is:
ΔKE = ΔPE
Now, the initial kinetic energy is zero since the grindstone started from rest. So, ΔKE is just the final kinetic energy.
ΔKE = 1/2 * I * ω^2
Where I is the moment of inertia and ω is the angular speed.
We can also calculate the change in potential energy, which is given by the work done by the torque:
ΔPE = Torque * Δθ
Where Δθ is the change in angle in radians.
Now, let's substitute these equations in the energy principle equation:
1/2 * I * ω^2 = Torque * Δθ
Plugging in the given values:
1/2 * 0.168 kg·m^2 * ω^2 = 22.9 N·m * 17.1 rev
Now, let's do some math to solve for ω. Trust me, it's an integral part of clown calculations!
*clownishly scribbles on a calculator*
Calculating, calculating... and the answer is...
ω ≈ 10.53 rev/s
Ta-da! The angular speed, after the grindstone has made 17.1 rev, is approximately 10.53 rev/s. Keep on grinding, my friend!
To solve this problem using energy principles, we can use the formula for rotational kinetic energy:
KE = 0.5 * moment of inertia * angular speed^2
Since the grindstone starts from rest, its initial angular speed (ω0) is 0.
The work done by the constant torque to rotate the grindstone can be calculated using the formula:
Work = torque * angular displacement
Here, the angular displacement is 17.1 revolutions, which can be converted to radians by multiplying by 2π (1 revolution = 2π radians).
So, the work done is:
Work = torque * (angular displacement in radians)
= 22.9 N·m * (17.1 rev * 2π rad/rev)
= 22.9 N·m * (17.1 * 2π) rad
Now, the work done is equal to the change in kinetic energy:
Work = KE - KE0
Since the initial kinetic energy (KE0) is 0, we can say:
KE = Work
Substituting the value of work:
0.5 * moment of inertia * angular speed^2 = 22.9 N·m * (17.1 * 2π) rad
Simplifying the equation:
angular speed^2 = (22.9 N·m * (17.1 * 2π) rad) / (0.5 * moment of inertia)
angular speed^2 = (22.9 N·m * 2π * 17.1 rad) / (0.168 kg·m^2)
angular speed^2 ≈ 1533.48 rev^2/s^2
Taking the square root of both sides to solve for the angular speed:
angular speed ≈ √(1533.48 rev^2/s^2)
Therefore, the angular speed after the grindstone has made 17.1 revolutions is approximately:
angular speed ≈ 39.16 rev/s
To find the angular speed of the grindstone after it has made 17.1 revolutions, we can use the principle of conservation of energy, assuming no friction is present.
The initial kinetic energy of the grindstone is zero because it starts from rest. The final kinetic energy can be calculated using the rotational kinetic energy formula:
K = (1/2) * I * ω^2
Where:
K is the kinetic energy,
I is the moment of inertia,
ω is the angular speed.
The torque applied to the grindstone, τ, is given as 22.9 N·m.
Torque is related to angular acceleration, α, using the equation:
τ = I * α
Rearranging the equation, we can find the angular acceleration:
α = τ / I
Now, using the formula for rotational kinematics, we can find the angular speed:
ω^2 = ω0^2 + 2αθ
Where:
ω0 is the initial angular speed (zero in this case),
θ is the angle of rotation (17.1 revolutions, which is equal to 2π * 17.1 radians).
Substituting the values and solving the equation:
ω^2 = 0 + 2 * (22.9 N·m / 0.168 kg·m^2) * (2π * 17.1 rad)
ω^2 = 2 * (22.9 / 0.168) * (2π * 17.1)
ω^2 = 2 * 318.8 * (2π * 17.1)
Simplifying:
ω^2 = 36389.248
Taking the square root of both sides:
ω ≈ 190.944 rev/s
Therefore, the angular speed of the grindstone after it has made 17.1 revolutions is approximately 190.944 rev/s.