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Unformatted text preview: homework 27 – BAUTISTA, ALDO – Due: Apr 5 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 3. 10 points. Consider an Earthlike planet hit by an asteroid. The planet has mass M p = 6 . 35 × 10 23 kg and radius R p = 7060000 m, and you may ap proximate it as a solid ball of uniform density. It rotates on its axis once every T = 39 hr. The asteroid has mass M a = 3 . 09 × 10 17 kg and speed v a = 35700 m / s (relative to the planet’s center); its velocity vector points θ = 74 ◦ below the Eastward horizontal. 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. Correct answer: 5 . 66573 × 10 32 kg m 2 / s (tol erance ± 1 %). Explanation: Basic Concept: A rigid body rotating around a fixed axis has angular momentum L = I × ω where I is the body’s moment of inertia about the axis of rotation. Approximating the planet as a solid ball of uniform density, its moment of inertia is I p = 2 5 M R 2 = 1 . 26603 × 10 37 kg m 2 . Its angular velocity before the impact is ω = 2 π T = 4 . 4752 × 10 5 rad / s , so L = I × ω = I × ω = (1 . 26603 × 10 37 kg m 2 ) × (4 . 4752 × 10 5 rad / s) = 5 . 66573 × 10 32 kg m 2 / s . Question 2 Part 2 of 3. 10 points. Calculate the asteroid’s angular momen tum relative to the planetary axis. Correct answer: 2 . 14669 × 10 28 kg m 2 / s (tol erance ± 1 %). Explanation: Approximating the asteroid as a pointlike particle, its angular momentum is ~ L a = ~ R × M a ~v a where ~ R is the asteroid’s radius vector and M a ~v a is its linear momentum vector. At the moment of impact, both the radius vector ~ R and the momentum vector M a ~v a of the asteroid lie in the planet’s equatorial plane. Consequently, their vector product is perpendicular to the equatorial plane and parallel to the planet axis. Because the hori zontal component of the asteroid’s velocity is directed Eastward — the same as the planet’s rotation — the asteroid’s angular momentum ~ L a has the same direction as the planet’s an gular momentum ~ L p . In magnitude, L a = R p M a × horizontal component of v a = R p M a × v a cos θ = (7060000 m) × (3 . 09 × 10 17 kg) (35700 m / s) cos(74 ◦ ) = 2 . 14669 × 10 28 kg m 2 / s ....
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 Spring '08
 Turner
 Angular Momentum, Mass, Work, Rigid Body, Rotation, Correct Answer, BAUTISTA

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