{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Hmwk - Alvarado Patrick Homework 6 Due 10:00 am Inst Andrei...

This preview shows pages 1–3. Sign up to view the full content.

Alvarado, Patrick – Homework 6 – Due: Oct 19 2006, 10:00 am – Inst: Andrei Sirenko 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points A cylinder with moment of inertia 23 . 6 kg m 2 rotates with angular velocity 5 . 5 rad / s on a frictionless vertical axle. A second cylinder, with moment of inertia 40 . 4 kg m 2 , initially not rotating, drops onto the first cylinder and remains in contact. Since the surfaces are rough, the two eventually reach the same an- gular velocity. I 2 I 1 ω 0 Before ω After Calculate the final angular velocity. Correct answer: 2 . 02813 rad / s. Explanation: From conservation of angular momentum ( I 1 + I 2 ) ω = I 1 ω 0 , or ω = I 1 I 1 + I 2 ω 0 = (23 . 6 kg m 2 ) (23 . 6 kg m 2 ) + (40 . 4 kg m 2 ) (5 . 5 rad / s) = 2 . 02813 rad / s . 002 (part 2 of 2) 10 points Show that energy is lost in this situation by calculating the ratio of the final to the initial kinetic energy. Correct answer: 0 . 36875 . Explanation: K i = 1 2 I 1 ω 0 2 and K f = 1 2 ( I 1 + I 2 ) ω 2 , Therefore K f = 1 2 ( I 1 + I 2 ) ω 2 = 1 2 ( I 1 + I 2 ) I 1 I 1 + I 2 2 ω 2 0 = I 1 I 1 + I 2 1 2 I 1 ω 2 0 = I 1 I 1 + I 2 K i . So K f K i = I 1 I 1 + I 2 = (23 . 6 kg m 2 ) (23 . 6 kg m 2 ) + (40 . 4 kg m 2 ) = 0 . 36875 < 1 . keywords: 003 (part 1 of 1) 10 points Two disks of identical mass but different radii ( r and 2 r ) are spinning on frictionless bear- ings at the same angular speed ω 0 but in op- posite directions. The two disks are brought slowly together. The resulting frictional force between the surfaces eventually brings them to a common angular velocity. 2 r ω 0 r ω 0 What is the magnitude of that final angular velocity in terms of ω 0 ?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Alvarado, Patrick – Homework 6 – Due: Oct 19 2006, 10:00 am – Inst: Andrei Sirenko 2 1. ω f = 4 5 ω 0 2. ω f = 2 5 ω 0 3. ω f = 1 4 ω 0 4. ω f = 1 3 ω 0 5. ω f = 1 5 ω 0 6. ω f = 3 4 ω 0 7. ω f = 1 2 ω 0 8. ω f = 3 5 ω 0 correct 9. ω f = 2 3 ω 0 Explanation: Note: Since the disks are spinning in oppo- site directions, let ω 1 = ω 0 and ω 2 = - ω 0 . The inertia of the larger disk is I 1 = 1 2 m (2 r ) 2 = 2 m r 2 , and of the smaller disk I 2 = 1 2 m r 2 . Using conservation of angular momentum, I i ω i = I f ω f I 1 ω 0 - I 2 ω 0 = ( I 1 + I 2 ) ω f ω f = I 1 - I 2 I 1 + I 2 ω 0 = 2 m r 2 - 1 2 m r 2 2 m r 2 + 1 2 m r 2 ω 0 = 3 5 ω 0 . keywords: 004 (part 1 of 3) 10 points Consider an Earth-like planet hit by an aster- oid. The planet has mass M p = 5 . 54 × 10 23 kg and radius R p = 5 . 89 × 10 6 m, and you may approximate it as a solid ball of uniform density. It rotates on its axis once every T = 15 hr. The asteroid has mass M a = 3 . 82 × 10 17 kg and speed v a = 23200 m / s (relative to the planet’s center); its velocity vector points θ = 73 below the Eastward hor- izontal. The impact happens at an equatorial location. The picture below shows the view from above the planet’s North pole: v m θ ω R First, calculate the planet’s angular mo- mentum (relative to its spin axis) before the impact.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

Hmwk - Alvarado Patrick Homework 6 Due 10:00 am Inst Andrei...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online