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Torque=momentinertial*angular acceleration
solve for angular acceleration.
A torque of 1.15 N·m is applied to a bicycle wheel of radius 36 cm and mass 0.83 kg. Treating the wheel as a hoop, find its angular acceleration.
?rad/s2
I know accel is alpha times the radius...where does torque tie in?
bobpursley
answered16 years ago
Explain Bot
answered11 months ago
To find the angular acceleration of the bicycle wheel, we need to consider the torque applied to it. Torque is the force applied to an object multiplied by the distance from the point of rotation (axis of rotation). In mathematical terms, torque (τ) = force (F) * lever arm (r), where τ=F*r.
In this case, the torque applied to the bicycle wheel is given as 1.15 N·m. The lever arm, or radius of the wheel, is 36 cm, which should be converted to meters by dividing it by 100 (as 1 m = 100 cm).
So, the torque (τ) is 1.15 N·m, and the radius (r) is 36 cm / 100 = 0.36 m.
Now, we can use Newton's second law for rotational motion, which states that torque (τ) equals the moment of inertia (I) multiplied by the angular acceleration (α). In mathematical terms, τ = I * α.
For a hoop-shaped object like a bicycle wheel, the moment of inertia (I) is given by the formula I = m * r², where m is the mass of the wheel and r is the radius.
Let's calculate the moment of inertia. The mass of the wheel is given as 0.83 kg, and the radius is 0.36 m.
I = 0.83 kg * (0.36 m)² = 0.1075 kg·m²
Now we have all the necessary values, so we can rearrange the torque equation to solve for angular acceleration (α).
τ = I * α
Rearranging,
α = τ / I
α = (1.15 N·m) / (0.1075 kg·m²)
Evaluating the expression,
α ≈ 10.70 rad/s²
Therefore, the angular acceleration of the bicycle wheel is approximately 10.70 rad/s².