Physics for Scientists and Engineers, Vol. 1
Physics for Scientists and Engineers, Vol. 1
6th Edition
ISBN: 9781429201322
Author: Paul A. Tipler, Gene Mosca
Publisher: Macmillan Higher Education
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Chapter 10, Problem 74P

(a)

To determine

To Calculate:

The horizental component of angular momentum and the vertical componeent of angular momentum.

(a)

Expert Solution
Check Mark

Answer to Problem 74P

  Lx=[(5.36)(cosωti^+sinωtj^)]JsLy=(3.09k^)Js

Explanation of Solution

Given data:

Mass of the ball is m

Formula Used:

Angular momentum: L=r×p

Where, p is the linear momentum and r is the distance to rotational axis.

Newton’s second law of motion:

  F=ma

Where, m is the mass and a is the acceleration.

  

Calculation:

  Physics for Scientists and Engineers, Vol. 1, Chapter 10, Problem 74P

By the expression of the angular momentum of the ball about the point of support as

  L=r×p=r×mv=m(r×v).........(1)

Then, by applying Newton’s second law to the ball to get

  F(x)=Tsinθ=mv2rsinθ.........(2)

And F(z)=Tcosθmg=0.........(3)

Now, eliminate T fromequation (3):

  T=mgcosθ

By substituting the T value in equation (2).

  (mgcosθ)sinθ=mv2rsinθ

Then solve for the velocity of the ball

  v=rgsinθtanθ

Substitute numerical values to get

  v=rgsinθtanθ=(1.5)(9.8)sin30otan30o=2.06m/s

Express the position vector of the ball as

  r=(1.5m)sin30o(cosωti^+sinωtj^)(1.5m)cos30ok^

Find the velocity of the ball:

  v=drdt=(0.75ω)(sinωti^+cosωtj^)

Evaluate for ω :

  ω=2.06m/s(1.5m)sin30o=2.75rad/s

Substitute for ω to obtain

  v=(2.06m/s)(sinωti^+cosωtj^)

Substitute into equation (1) and evaluate L to get

  L=(2kg)[(1.5m)sino(cosωti^+sinωtj^)(1.5m)cos30ok^]×(2.06m/s)(sinωt^+cosωtj^)=[(5.36)(cosωti^+sinωtj^)+3.09k^]kgm2/s

Therefore, the horizontal component of angular momentum L is

  Lx=[(5.36)(cosωti^+sinωtj^)]kgm2/s

And the vertical component of angular momentum L is

  Ly=(3.09k^)kgm2/s

Conclusion:

  Lx=[(5.36)(cosωti^+sinωtj^)]kgm2/sLy=(3.09k^)kgm2/s

(b)

To determine

To Calculate:The magnitude of the torque exerted by gravity about the point of support.

(b)

Expert Solution
Check Mark

Answer to Problem 74P

  14.7 Nm

Explanation of Solution

Given data:

From the previous part:

  Lx=[(5.36)(cosωti^+sinωtj^)]kgm2/sLy=(3.09k^)kgm2/s

Formula Used:

The magnitude of the torque exerted by gravity about the point of support as

  τ=mgrsinθ

Calculation:

By differentiating the angular momentum:

  dLdt=ddt[(5.36)(cosωti^+sinωtj^)+3.09k^]=(5.36)(sinωti^+cosωtj^)Nm

Therefore, the magnitude of above derivative

  |dLdt|=(5.36)(2.75)=14.7Nm

Substitute the values:

  τ=mgrsinθ=(2kg)(9.81m/s2)(1.5m)sin30o=14.7Nm

Conclusion:

Torque exerted by gravity about the point of support is 14.7 Nm .

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