Cross Product in Uniform Circular Motion

In summary, the conversation discusses a problem involving a centripetal-acceleration addict riding in uniform circular motion. The question asks for the values of r X a at a specific instant, but the student is unsure of how to approach the problem. They know that acceleration is always towards the center and perpendicular to the object's velocity, and that a cross product of two vectors pointing in opposite directions would result in 0. The student is advised to approach the problem through components or the formula for the cross product.
  • #1
snoggerT
186
0

Homework Statement



A centripetal-acceleration addict rides in uniform circular motion with period T = 3.22 s and radius r = 3.00 m. At one instant his acceleration is a = (7.00 m/s2) i + (-9.00 m/s2) j. At that instant, what are the following values?

(b) r X a





The Attempt at a Solution



I'm really stumped on this one. I completely understand how to do cross products of a vector, but there is nothing in my textbook about doing a cross product problem in uniform circular motion. I tried taking the magnitude of a and multiplying it to times r, but that wasn't right. If anyone can give me a hint as to how to start this, I would be very thankful.
 
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  • #2
snoggerT said:

Homework Statement



A centripetal-acceleration addict rides in uniform circular motion with period T = 3.22 s and radius r = 3.00 m. At one instant his acceleration is a = (7.00 m/s2) i + (-9.00 m/s2) j. At that instant, what are the following values?

(b) r X a





The Attempt at a Solution



I'm really stumped on this one. I completely understand how to do cross products of a vector, but there is nothing in my textbook about doing a cross product problem in uniform circular motion. I tried taking the magnitude of a and multiplying it to times r, but that wasn't right. If anyone can give me a hint as to how to start this, I would be very thankful.

what do you know about the acceleration vector in uniform circular motion?
 
  • #3
nrqed said:
what do you know about the acceleration vector in uniform circular motion?

- I know that acceleration is always concentrated toward the center and is always perpendicular to the objects velocity. I know that a=v^2/r . I think that's about it...
 
  • #4
snoggerT said:
- I know that acceleration is always concentrated toward the center and is always perpendicular to the objects velocity. I know that a=v^2/r . I think that's about it...

Right. So in the formula [itex] \vec{r} \times \vec{a} [/itex], assuming that by the vector "r" they mean the vector pointing from the center of the circle to the position of the object, you know that these two vectors point in opposite directions. What can you say about the cross product of two vectors pointing in opposite directions?
 
  • #5
nrqed said:
Right. So in the formula [itex] \vec{r} \times \vec{a} [/itex], assuming that by the vector "r" they mean the vector pointing from the center of the circle to the position of the object, you know that these two vectors point in opposite directions. What can you say about the cross product of two vectors pointing in opposite directions?

that their product would be 0. Thanks for the help. I just wish the textbook explained things worth a crap.
 
  • #6
snoggerT said:
that their product would be 0. Thanks for the help. I just wish the textbook explained things worth a crap.

You're welcome. When you have questions about the cross product, there are two ways to go: through components or through the formula "magnitude of a times magnitude of b times sin (theta)".
Then think about all the information that you know about either the magnitudes of the vectors or their direction.


best luck!
 

Related to Cross Product in Uniform Circular Motion

What is the cross product in uniform circular motion?

The cross product in uniform circular motion is a mathematical operation that calculates the vector perpendicular to the plane in which an object is moving. It is used to determine the direction of the angular velocity of an object in circular motion.

How is the cross product calculated?

The cross product is calculated by multiplying the magnitude of two vectors by the sine of the angle between them. This results in a new vector that is perpendicular to both of the original vectors.

What is the significance of the cross product in uniform circular motion?

The cross product is significant in uniform circular motion because it helps us understand the relationship between the angular velocity and linear velocity of an object. It also helps us determine the direction of the centripetal force acting on the object.

Can the cross product be negative?

Yes, the cross product can be negative. The sign of the cross product is determined by the right-hand rule, where the direction of the resulting vector is perpendicular to the plane in which the two original vectors lie. If the resulting vector points in the opposite direction, it is considered a negative cross product.

How is the cross product related to torque in uniform circular motion?

In uniform circular motion, the cross product is directly related to torque. The magnitude of the torque is equal to the cross product of the force vector and the vector from the pivot point to the point of application of the force. This relationship is described by the equation τ = r x F.

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