Senin, 27 Oktober 2008

Problems Impulse and Momentum

i. I was wondering if anyone could help me with this problem.

A baseball of mass 850 g travelling at 15 m/s due south is caught and thrown northwest at the same speed during a time interval of 0.85 s.
(a) What is the magnitude of the change in the momentum of the ball during this time interval?
(b) What is the direction of the change in the ball's momentum during this time interval?
(c) What are the magnitude and the direction of the impulse delivered to the ball during this time interval?

There are more questions, but I know how to do those.
I have solved the problem, but I got different answers from the answer key. I don't know what I am missing.

(a) I got this one correct 24 kg*m/s.
(b) I got 22.5° East of South. They say it is 135°, no direction specified. 135° is the angle between the two vectors. Why would that be the direction of the change in the momentum?
(c) I got 24 N*s 22.5° East of South. They say it is 24N*s 22.5° West of North. Why is the direction of the change in momentum different from the direction of the impulse? Why do I have a direction of East of South and they say it is West of North?

Any help would be greatly appreciated. Thank you.

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TORQUE

torque (τ) in physics, also called a moment (of force), is a pseudo-vector that measures the tendency of a force to rotate an object about some axis[1] (center). The magnitude of a torque is defined as the product of the component of the force perpendicular to the length of the lever arm[2] (radius). Just as a force is a push or a pull, a torque can be thought of as a twist.

The SI unit for torque is the newton meter (N m). In Imperial and U.S. customary units, it is measured in foot pounds (ft·lbf) (also known as 'pound feet') and for smaller measurement of torque: inch pounds (in·lbf) or even inch ounces (in·ozf) . The symbol for torque is τ, the Greek letter tau. .

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[edit] History

The concept of torque, also called moment or couple, originated with the studies of Archimedes on levers. The rotational analogues of force, mass, and acceleration are torque, moment of inertia, and angular acceleration, respectively.

[edit] Explanation

The force applied to a lever multiplied by its distance from the lever's fulcrum, the length of the lever arm, is its torque. A force of three newtons applied two meters from the fulcrum, for example, exerts the same torque as one newton applied six meters from the fulcrum. This assumes the force is in a direction at right angles to the straight lever. The direction of the torque can be determined by using the right hand grip rule: curl the fingers of your right hand to indicate the direction of rotation, and stick your thumb out so it is aligned with the axis of rotation. Your thumb points in the direction of the torque vector.[3]

Mathematically, the torque on a particle (which has the position r in some reference frame) can be defined as the cross product:

\mathbf{\tau} = \mathbf{r} \times \mathbf{F}

where

r is the particle's position vector relative to the fulcrum
F is the force acting on the particle.

The torque on a body determines the rate of change of its angular momentum,

\mathbf{\tau}=\frac{\mathrm{d}\mathbf{L}}{\mathrm{d}t}

where

L is the angular momentum vector
t is time.

As can be seen from either of these relationships, torque is a vector, which points along the axis of the rotation it would tend to cause.

[edit] Proof of the equivalence of definitions

The definition of angular momentum for a single particle is:

\mathbf{L} = \mathbf{r} \times \mathbf{p}

where "×" indicates the vector cross product. The time-derivative of this is:

\frac{d\mathbf{L}}{dt} = \mathbf{r} \times \frac{d\mathbf{p}}{dt} + \frac{d\mathbf{r}}{dt} \times \mathbf{p}

This result can easily be proven by splitting the vectors into components and applying the product rule. Now using the definitions of velocity v = dr/dt, acceleration a = dv/dt and linear momentum p = mv, we can see that:

\frac{d\mathbf{L}}{dt} = \mathbf{r} \times m \frac{d\mathbf{v}}{dt} +  \mathbf{v} \times m\mathbf{v}