# Torque Self-Test: Angular Acceleration

You spin a bicycle wheel (diameter of 0.85 m, mass of 4.5 kg), applying a force of 24 N tangentially. You will find the angular acceleration of the wheel by the following steps.

(a) What is the torque on the wheel?

• A. 5.1 N m
• B. 9.8 N m
• C. 10.2 N m
• D. 20.4 Nm

• A. No. You have made too many divisions by 2
• B. No Weight is not involved
• C. Correct
• D. You have probably used the diameter instead of the radius.

(b) Assuming the wheel is a thin-walled hollow cylinder, what is its moment of inertia?

#### Moment of Inertia

Moment of inertia is the rotational analogue to mass.
Here is a list of moments of inertia for various bodies:

Shape Axis Equation
slender Rod axis through center $I = (1/12)ML^2$
slender Rod axis through end $I = (1/3)ML^2$
rectangular plane axis through center $I = (1/2)M(a^2 + b^2)$
rectangular plane axis along edge $I = (1/3)Ma^2$
cylinder hollow $I = (1/2)M(R_{1^2} + R_{2^2})$
cylinder solid $I = (1/2)MR^2$
cylinder thin-walled hollow $I = MR^2$
sphere solid $I = (2/5)MR^2$
sphere thin-walled hollow $I = (2/3)MR^2$

• A. 0.761 kg m2
• B. 0.813 kg m2
• C. 1.015 kg m2
• D. 1.137 kg m2

• A. No: Remember it is a THIN cylinder (or ring)
• B. Correct
• C. No: Remember it is a THIN cylinder (or ring)
• D. No: Remember it is a THIN cylinder (or ring)

(c) Find the angular acceleration, $\alpha$ , of the wheel.

• A. Correct
• B. No. Remember for this wheel tou have determined that $I = 0.813 Nm^2$ and Torque  $= 10.2 Nm$
• C.  No. Remember for this wheel tou have determined that $I = 0.813 Nm^2$ and Torque  $= 10.2 Nm$
• D.  No. Remember for this wheel tou have determined that $I = 0.813 Nm^2$ and Torque  $= 10.2 Nm$

#### Solutions

(a) The torque is:

$\tau = r \times F$

$= (0.425) \cdot (24) \cdot \sin (90)$

$= 10.2 N \cdot m$

(b) The moment of inertia for a thin-walled, hollow cylinder is:

$I = M \cdot R^2$

$= (4.5) \cdot 0.425^2$

$= 0.813 kg\cdot m^2$

(c) Recall that the net torque is equal to the moment of inertia multiplied by angular acceleration:

$\sum \tau - I \cdot \alpha$

Since there is only one torque on the bicycle wheel, then the net torque is simply $\tau$:

$\tau = I \cdot \alpha$

Rearranging the above equation for $\alpha$ , and substituting for $\tau$ and $I$, we get:

$\alpha = \frac {T}{I} = \frac{10.2}{0.813} = 12.5 s^{-2}$