A wheel of radius 0.398 m is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about the axis is 0.0576 kg·m2. A massless cord wrapped around the wheel is attached to a 0.790 kg block that slides on a horizontal frictionless surface. If a horizontal force of magnitude P = 3.52 N is applied to the block , what is the angular acceleration of the wheel? Take the clockwise direction to be the negative direction and assume the string does not slip on the wheel.
ma=P-T
Iε=M=TR
T=P-ma=P-mεR,
T=Iε/R.
P-mεR= Iε/R.
ε=PR/(I+mR²)=
=3.52•0.398/(0.0576+0.790•0.398²)=7.67 rad/s².
To find the angular acceleration of the wheel, we can use the following equation:
τ = Iα
Where:
- τ is the torque applied to the wheel
- I is the rotational inertia of the wheel
- α is the angular acceleration of the wheel
Now, let's calculate the torque applied to the wheel:
τ = r * F
Where:
- r is the radius of the wheel
- F is the force applied to the block
Substituting the values given:
τ = (0.398 m) * (3.52 N)
τ = 1.3976 N·m
Now we can rearrange the torque equation to solve for the angular acceleration:
α = τ / I
Substituting the values:
α = (1.3976 N·m) / (0.0576 kg·m^2)
α ≈ 24.25 rad/s^2
Therefore, the angular acceleration of the wheel is approximately 24.25 rad/s^2.
To find the angular acceleration of the wheel, we can use the following equation:
τ = I * α
Where:
τ is the torque applied to the wheel
I is the rotational inertia of the wheel
α is the angular acceleration of the wheel
First, we need to find the torque exerted on the wheel by the applied force.
The torque can be calculated using the formula:
τ = r * F * sin(θ)
Where:
r is the radius of the wheel
F is the applied force
θ is the angle between the force and the radius vector of the wheel (which is 90 degrees in this case)
Substituting the given values:
r = 0.398 m
F = 3.52 N
θ = 90 degrees = π/2 radians
τ = (0.398 m) * (3.52 N) * sin(π/2)
τ = 1.392 N·m
Now, we can use the torque and the rotational inertia of the wheel to find the angular acceleration:
τ = I * α
1.392 N·m = (0.0576 kg·m^2) * α
Solving for α:
α = (1.392 N·m) / (0.0576 kg·m^2)
α ≈ 24.17 rad/s^2
So, the angular acceleration of the wheel is approximately 24.17 rad/s^2.