A constant force of 40 N is applied tangentially to the rim of a wheel with 20 cm radius.
The wheel has a moment of inertia of 30 kg m2.
Show that the angular acceleration of
the wheel is 0.27 rad s-2
oo...i totally forgot to expand torque to be the force x perpend. radius
thanks..got it
To find the angular acceleration of the wheel, we can use Newton's second law for rotational motion, which states that the torque applied to an object is equal to the moment of inertia multiplied by the angular acceleration.
The torque formula is given by:
τ = I α
where:
τ = torque,
I = moment of inertia, and
α = angular acceleration.
In this case, the torque (τ) is equal to the product of the applied force (F) and the radius of the wheel (r). So, we have:
τ = F × r
Substituting the given values, we get:
τ = 40 N × 0.2 m = 8 N∙m
Now, we can rearrange the torque formula to solve for the angular acceleration (α):
α = τ / I
Substituting the given moment of inertia (I = 30 kg∙m^2), we get:
α = 8 N∙m / 30 kg∙m^2
Simplifying this expression, we find:
α ≈ 0.27 rad/s^2
Therefore, the angular acceleration of the wheel is approximately 0.27 rad/s^2.
To find the angular acceleration of the wheel, we can use the equation:
τ = I * α
where τ is the torque applied to the wheel, I is the moment of inertia of the wheel, and α is the angular acceleration of the wheel.
In this case, the torque applied to the wheel is given by the formula:
τ = F * r
where F is the force applied tangentially to the rim of the wheel, and r is the radius of the wheel.
Given that F = 40 N and r = 20 cm = 0.2 m, we can substitute these values into the torque equation:
τ = 40 N * 0.2 m = 8 Nm
Now, substituting the value of τ into the first equation:
8 Nm = 30 kg m^2 * α
Solving for α:
α = 8 Nm / 30 kg m^2 = 0.27 rad s^-2
Therefore, the angular acceleration of the wheel is 0.27 rad s^-2.
Torque=I*alpha
force*radius=I(alpha
solve for alpha.