# 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 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 m
^{2} - B. 0.813 kg m
^{2} - C. 1.015 kg m
^{2} - D. 1.137 kg m
^{2}

- 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. 12.5 rad/s
^{2} - B. 14.0 rad/s
^{2} - C. 15.8 rad/s
^{2} - D. 18.4 rad/s
^{2}

**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:

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