A cord is wrapped around the rim of a wheel .250 m in radius, and a steady pull of 40.0 N is exerted on the cord. The wheel is mounted on frictionless bearings on a horizontal shaft through its center. The moment of inertia of the wheel about this shaft is 5.00kg*m^2. Compute the angular acceleration of the wheel.

The torque is T = 40 N x 0.25 M = 10 N-m
The moment of inertia is I = 5.00 kg/m^2
The angular acceleration in radians/s^2
alpha = T/I
Do the numbers

The torque, T, is calculated by multiplying the force exerted on the cord, 40.0 N, by the radius of the wheel, 0.25 m.

T = 40.0 N * 0.25 m = 10.0 N-m

The moment of inertia, I, is given as 5.00 kg*m^2.

Now, we can calculate the angular acceleration, α, using the formula:

α = T / I

Substituting in the values,

α = 10.0 N-m / 5.00 kg*m^2

α = 2.00 radians/s^2

Therefore, the angular acceleration of the wheel is 2.00 radians/s^2.

The torque is calculated by multiplying the applied force (40.0 N) by the distance from the axis of rotation (0.250 m). So, T = 40.0 N x 0.250 m = 10.0 N-m.

The moment of inertia (I) of the wheel about the axis of rotation is given as 5.00 kg*m^2.

To compute the angular acceleration (alpha) of the wheel, we can use the equation alpha = T/I.

Substituting the given values, we have:

alpha = 10.0 N-m / 5.00 kg*m^2

Calculating the division, we find:

alpha = 2.00 (N-m) / (kg*m^2)

So, the angular acceleration of the wheel is 2.00 rad/s^2.

Please note that when calculating the numerical answer, it is important to ensure that the units are consistent throughout the calculations.

R1R2F/(I+mR2square)

2.00