Centripetal acceleration problem

[RadiationX]

I have the solution to this problem but I'm not understanding the concept behind the solution.

An object moves at constant speed along a circular path in a horizontal xy plane,with the center at the origin.When the object is at x=-2m its velocity is -(4m/s) in the J (hat) direction. Give the objects velocity and acceleration when it is at y=2m.


the answers -(4ms) in the i (hat) direction and -(8ms^s) in the j (hat direction)

 

[dextercioby]

I really don't see a way to find the centripetal acceleration.You'd basically need "omega".What about the answer for the velocity...?To me,it seems 2 as big as it should be.

Daniel.

 

[xanthym]

I have the solution to this problem but I'm not understanding the concept behind the solution.

An object moves at constant speed along a circular path in a horizontal xy plane,with the center at the origin.When the object is at x=-2m its velocity is -(4m/s) in the J (hat) direction. Give the objects velocity and acceleration when it is at y=2m.


the answers -(4ms) in the i (hat) direction and -(8ms^s) in the j (hat direction)

For the given uniform circular motion around the origin, the velocity "v" will have constant magnitude and be perpendicular to a radius drawn from the origin (center) to the object's position on the circle. When the object's velocity "v" is (-4 m/s)j at x=(-2 m), the radius from origin to object will lie along the x-axis, and therefore the object must be crossing the negative x-axis x=(-2 m). Thus, the circle radius is r=(2 m), and the object is moving counter-clockwise.

When the object is at y=(+2 m), it will be crossing the positive y-axis (since the circle's radius is r=2) in the counter-clockwise direction with the same magnitude as before. Thus, its velocity here is v=(-4 m/s)i. The acceleration "A" at this point must be perpendicular to its path and directed towards the circle's center. Hence, it will have direction (-1)j and magnitude (v^2)/r. Using the values for |v|=(4) and r=(2) yields A=(-8 m/sec^2)j.
~

 

[Kelvin]

I have the solution to this problem but I'm not understanding the concept behind the solution.

An object moves at constant speed along a circular path in a horizontal xy plane,with the center at the origin.When the object is at x=-2m its velocity is -(4m/s) in the J (hat) direction. Give the objects velocity and acceleration when it is at y=2m.


the answers -(4ms) in the i (hat) direction and -(8ms^s) in the j (hat direction)

it is uniform circular motion... at any time the speed is constant

at x = -2 m the object is moving in -j direction only, and we immediately know that the object travels with speed 4 m/s in a circular path of radius 2 m in anti-clockwise direction.

so at y = 2 m, the x coordinate should be zero. and the velocity should be in -i direction. since there's not j component, v = -4 m/s i

and you just substitude numbers in the formula you should get the centripetal acceration

 

[ehild]

I have the solution to this problem but I'm not understanding the concept behind the solution.

An object moves at constant speed along a circular path in a horizontal xy plane,with the center at the origin.When the object is at x=-2m its velocity is -(4m/s) in the J (hat) direction. Give the objects velocity and acceleration when it is at y=2m.


the answers -(4ms) in the i (hat) direction and -(8ms^s) in the j (hat direction)

By = - 2 m you mean the point (-2,0), don't you? If so, this point belongs to the circle the object moves along (see attached picture) and this means that the radius of the circle is r = 2 m. It is also clear that the object moves anti-clockwise. From the relation between the linear velocity and angular velocity you get \omega = 2 1/s. The linear velocity is perpendicular to the radius, the speed is constant, so the velocity at P' ( 0,2) is
\vec{v} = -4 \vec {i}
and the centripetal acceleration is
\vec {a} = - \omega^2 r \vec {j} = -8 \vec {j}

ehild

 

[RadiationX]

thanks i got it