A torque of 0.97 N-m is applied to a bicycle wheel of radius 0.35 m and mass 0.75 kg. Treating the wheel as a hoop, find its angular acceleration.
To find the angular acceleration of the bicycle wheel, we can use the equation:
τ = Iα
Where:
τ is the torque applied to the wheel,
I is the moment of inertia of the wheel,
and α is the angular acceleration.
In this case, the torque (τ) is given as 0.97 N-m.
The moment of inertia (I) for a hoop can be calculated as:
I = MR²
Where:
M is the mass of the wheel,
R is the radius of the wheel.
In this case, the mass (M) of the wheel is given as 0.75 kg and the radius (R) is given as 0.35 m.
Substituting these values into the above equation, we get:
I = (0.75 kg) x (0.35 m)²
Calculating this gives:
I = 0.091 kg·m²
Now we can solve for α by rearranging the equation:
α = τ / I
Plugging in the values we have:
α = 0.97 N-m / 0.091 kg·m²
Calculating this gives:
α ≈ 10.66 rad/s²
Therefore, the angular acceleration of the bicycle wheel is approximately 10.66 rad/s².
Look up the moment of inertia for a hoop.
torque=momeninertia)angular acceleration