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19 Nov 2019

For no apparent reason, a poodle is running at a constant speed of3.40 m/s in a circle with radius 2.9m. Let v1 be the velocityvector at time t1, and let v2 be the velocity vector at timet2

For change in t1 = 0.6s calculate the magnitude (to foursignificant figures) of the average acceleration?

For change in t1 = 0.6s calculate the direction (relative to v1) ofthe average acceleration, A(a/v)?

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19 Aug 2019

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

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70. A 4 µF capacitor is charged to a potential differ¬ence of 100V. The electrical energy stored in the capacitor is
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which of the vectors below best represents the direction of the impulse vector j⃗ ?

 

To learn about the impulse-momentum theorem and its applications in some common cases.

Using the concept of momentum, Newton's second law can be rewritten as

?F? =dp? dt, (1)

where ?F?  is the net force F? net acting on the object, and dp? dt is the rate at which the object's momentum is changing.

If the object is observed during an interval of time between times t1 and t2, then integration of both sides of equation (1) gives

?t2t1?F? dt=?t2t1dp? dtdt. (2)

The right side of equation (2) is simply the change in the object's momentum p2??p1?. The left side is called the impulse of the net force and is denoted by J? . Then equation (2) can be rewritten as

J? =p2??p1?.

This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force F? net acting along the direction of motion, the impulse-momentum theorem can be written as

F(t2?t1)=mv2?mv1. (3)

Here F, v1, and v2 are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately.

( The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass m moving along the x-axis. The net force F is acting on the particle along the x-axis. F is a constant force.)

Learning Goal: To learn about the impulse-momentum

A) The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0.

B) The particle starts from rest at t=0. What is the magnitude v of the velocity of the particle at time t? Assume that t>0.

C) The particle has the momentum of magnitude p1 at a certain instant. What is p2, the magnitude of its momentum ?t seconds later?

D) The particle has the momentum of magnitude p1 at a certain instant. What is v2, the magnitude of its velocity ?t seconds later?

E) Find the magnitude of the impulse J delivered to the particle.

F) Which of the vectors below best represents the direction of the impulse vector J??
(Figure 1)

1
2
3
4
5
6
7
8

 

 

 
 

(A) Show that the position of a particle on a circle of radius Rwith center at the origin (x=0,y=0) is given in unit vectors byr=cosθi+sinθj, where θ is the angle of the position vector with thex axis.

B) If the particle moves with constant speed v and period T,starting on the x axis at t=0, find an expression for θ in terms ofthe time and T.

C) Differentiate the position vector twice with respect to timeto find the acceleration and show that it is the centripetalacceleration, whose magnitude is given by equation (a=(v^2)/r) andwhose direction is toward the center of the circle. (Not that theunit vectors i and j are constants, independent of time.)

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