Explanation Given information: The path of particle is defined as, y = x 2 4 r 0 ε = 1 Equation 12.35 is defined as, d 2 u d θ 2 + u = F m h 2 u 2 Where, u = 1 r Equation 12.27 is defined as, 1 r = G M h 2 ( 1 + ε cos θ ) Calculation: At the given instant, y = x 2 4 r 0 r 0 − r cos θ = r 2 sin 2 θ 4 r 0 r 2 sin 2 θ + 4 r 0 r cos θ − 4 r 0 2 = 0 By solving, we get r = 2 r 0 ( 1 − cos θ ) sin 2 θ = 2 r 0 ( 1 − cos θ ) 1 − cos 2 θ = 2 r 0 1 + cos θ According to the above explanation, u = 1 r Then, u = 1 r = 1 + cos θ 2 r 0 Differentiate, d u d θ = − sin θ 2 r 0 d 2 u d θ 2 = − cos θ 2 r 0 According to equation 12

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

See similar textbooks

Concept explainers

Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be create…

Videos

Question

Chapter 12.3, Problem 12.97P

To determine

Using equation 12.35 and 12.37 that the central force F is inversely proportional to r2 and h=2GMr0.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 12.3, Problem 12.96P

Chapter 12.3, Problem 12.98P

Students have asked these similar questions

Three identical point masses of m = 0.1 (kg) are moving at a constant velocity v=10 (m/s) equidistant from each other on a circular orbit of radius R = 0.4 (m). What is the total angular momentum (kg.m²/s) of the three point masses relative to point A at the moment shown in the figure. Point A is 2R away from the center. m 2R A m R m A) 0.9 В) 1.2 С) 1.8 D) 3.0 E) 3.6

1. A wood block of mass mw = 200 g is projected from the ground at a speed vo = 35 m/s and at an angle 60 = 50°. When at height h= 24 m the wood block was hit by a bullet of mass m = 25 g which was moving at v= 220 m/s and at an angle 6 = 70°. After the collision, the bullet embedded itself in the block and move together and land at point A on the ground. Then they move on the ground that has coefficient of kinetic friction u: = 0.5 and start to compress a horizontal spring at point B. The distance between point À and B is 18 m and the spring constant k = 3x10? N/m. Right end of the spring is fixed, and all motions are confined in x-y plane. Wood block + Bullet Bullet Wood B block (a) Was the collision between the wood block and the bullet elastic or inelastic? Explain quantitively. (b) Find the maximum height from the ground the block with the bullet reached. (c) How far is point A from the launch point of the wood block? (d) Calculate the velocity at which the block-bullet system…

Q.1. A block of mass m = 0.5 kg is pushed against a horizontal spring of spring constant k = 450 N/m and negligible mass. The spring is compressed a distance x from equilibrium, and then released from rest. The block travels along a frictionless horizontal surface and reaches point B with a speed vg = 12 m/s. Take g = 10 m/s2. The compression of the spring is: a. x = 0.16 m b. x = 0.467 m c. x = 0.533 m B. d. x = 0.4 m

Chapter 12 Solutions

Vector Mechanics for Engineers: Dynamics

Ch. 12.1 - A 1000-Ib boulder B is resting on a 200-Ib...Ch. 12.1 - Marble A is placed in a hollow tube, and the tube...Ch. 12.1 - The two systems shown start from rest. On the...Ch. 12.1 - Prob. 12.CQ4P Ch. 12.1 - People sit on a Ferris wheel at points A, B, C,...Ch. 12.1 - Crate A is gently placed with zero initial...Ch. 12.1 - Prob. 12.F2P Ch. 12.1 - Objects A, B, and C have masses mA, mB, and...Ch. 12.1 - Blocks A and B have masses mAand mB, my...Ch. 12.1 - Blocks A and B have masses mAand mB, my...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Using equation 12.35 and 12.37 that the central force F is inversely proportional to r 2 and h = 2 G M r 0 .

Solution Summary: The author explains that the central force F is inversely proportional to r2.

Concept explainers

Videos

Using equation 12.35 and 12.37 that the central force F is inversely proportional to r2 and h=2GMr0.

Want to see the full answer?

Chapter 12 Solutions