## Relating Angular and Translational Quantities

10.3 Relating Angular and Translational Quantities

### Learning Objectives

By the end of this section, you will be able to:

- Given the linear kinematic equation, write the corresponding rotational kinematic equation
- Calculate the linear distances, velocities, and accelerations of points on a rotating system given the angular velocities and accelerations

Pause & Predict 10.3.1 |
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What is the moon’s acceleration as it orbits Earth? |

(a) 1.4 × 10^{-7} m/s^{2} |

(b) 2.7 × 10^{-6} m/s^{2} |

(c) 2.7 × 10^{-3} m/s^{2} |

(d) 2.0 × 10^{7} m/s^{2} |

(e) 5.2 × 10^{5} m/s^{2} |

Pause & Predict 10.3.2 |
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If the tangential acceleration is 100 m/s^{2}, what is the total acceleration of the yo-yo at point 2? |

(a) 780 m/s^{2} |

(b) 790 m/s^{2} |

(c) 880 m/s^{2} |

(d) 680 m/s^{2} |

(e) 300 m/s^{2} |

**Practice!**

Practice 10.3.1 |
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Earth is 149.6 billion meters from the Sun and takes 365 days to make one complete revolution around the Sun. Mars is 227.9 billion meters from the Sun and has an orbital period of 687 days. What is the ratio of Earth’s centripetal acceleration to Mars’s centripetal acceleration? |

(a) 1.88 |

(b) 0.809 |

(c) 1.24 |

(d) 0.430 |

(e) 2.33 |

Practice 10.3.2 |
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A 1000-kg car is moving at a constant speed around a circular turn with a radius of 18.5 meters. How fast must the car move to have an acceleration of 25.2 m/s^{2}? |

(a) 466 m/s |

(b) 21.6 m/s |

(c) 1.17 m/s |

(d) 34.3 m/s |

Practice 10.3.3 |
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A 1000-kg car is moving around a circular turn with a radius of 18.5 meters and decreasing in speed at a rate of 35.2 m/s^{2}. At the instant the car is moving at 16.8 m/s, what is the car’s total acceleration? |

(a) 15.3 m/s^{2} |

(b) 35.2 m/s^{2} |

(c) 50.5 m/s^{2} |

(d) 38.4 m/s^{2} |

(e) 19.9 m/s^{2} |

Practice 10.3.4 |
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A brother and sister are riding on a merry-go-round at the park. The brother rides on the outer edge of the merry-go-round and the sister rides closer to the center. While the merry-go-round rotates with a constant angular velocity, who has the greatest centripetal (radial) acceleration? |

(a) The brother has the greatest centripetal (radial) acceleration. |

(b) The sister has the greatest centripetal (radial) acceleration. |

(c) They both have the same non-zero centripetal (radial) acceleration. |

(d) They both have the same zero centripetal (radial) acceleration. |

While the merry-go-round rotates with a constant angular velocity, **the brother has the greatest centripetal (radial) acceleration**. The centripetal (radial) acceleration is calculated with where is the angular velocity and *r* is the radial distance from the center of the circular path, in this case the center of the merry-go-round. Since the brother is farther from the center than the sister and they have the same angular velocity, the brother’s centripetal acceleration is greater.

They cannot have the same centripetal acceleration because they are different distances from the center of the merry-go-round. They cannot have zero centripetal acceleration because their velocity vectors are changing direction throughout the motion and a change in the direction of a velocity vector results in a centripetal acceleration.

Practice 10.3.5 |
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A brother and sister are riding on a merry-go-round at the park. The brother rides on the outer edge of the merry-go-round and the sister rides closer to the center. While the merry-go-round rotates with an increasing angular velocity, who has the greatest tangential acceleration? |

(a) They both have the same non-zero tangential acceleration. |

(b) The sister has the greatest tangential acceleration. |

(c) The brother has the greatest tangential acceleration. |

(d) They both have the same zero tangential acceleration. |

While the merry-go-round rotates with an increasing angular velocity, **the brother has the greatest tangential acceleration**. Tangential acceleration is determined from the product of the angular acceleration and the radial distance from the center of the circular path: . Since the merry-go-round is increasing in angular velocity it has an angular acceleration, and this angular acceleration will be the same for the brother and the sister. Therefore, the brother, being farther away from the center of the merry-go-round, will have a greater tangential acceleration.

They cannot have the same tangential acceleration because they are different distances from the center of the merry-go-round. They cannot have zero tangential acceleration because they are increasing in angular velocity, which results in a tangential acceleration.

Practice 10.3.6 |
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A brother and sister are riding on a merry-go-round at the park. The brother rides on the outer edge of the merry-go-round and the sister rides closer to the center. While the merry-go-round rotates with an increasing angular velocity, who has the greatest linear acceleration? |

(a) The sister has the greatest linear acceleration. |

(b) The brother has the greatest linear acceleration. |

(c) They both have the same non-zero linear acceleration. |

(d) They both have the same zero tangential acceleration. |