Torque Self-Test: Angular Acceleration

• 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.

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

• 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)

• 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  $= 10.2 Nm$

(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:

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