Write Vector Expression in n-t and x-y coordinates of Acceleration

[Northbysouth]

Homework Statement

Write the vector expression for the acceleration a of the mass center G of the simple pendulum in both n-t and x-y coordinates for the instant when θ = 66° if θ'= 2.22 rad/sec and θ"= 4.475 rad/sec²

I have attached an image of the question.

Homework Equations

a_n = v²/r = rθ² = vθ'

a_t = v' = rθ'

The Attempt at a Solution

I've managed to calculate e_n and e_t correctly

a_t = (4.2 ft)(4.475 rad/sec²) = 18.795 ft/sec²

a_t = 18.795ft/sec²

For a_n I calculated velocity first:

v = rθ' = (4.2ft)(2.22 rad/sec)
v = 9.324 ft/sec

Hence a_n = (9.324 ft/sec)(2.22 rad/sec)
a_n = 20.69928 ft/sec²

Unfortunately, I'm now having difficulty with finding the velocity in terms of i and j.

I had thought that I could use geometry to do it:

a_tcos(90-66) = -17.77 i
a_tsin(90-66) = 7.644 j

But the system says it's wrong. Help is greatly appreciated.

 

[cosmic dust]

Why don't you try to express e_n and e_t in terms of e_x and e_y ? Doing that, you can find the total acceleration, a = a_n + a_n , in terms of its projections on x and y axis.* e _i is the unit vector in the direction of the subscript "i".

 

[SammyS]

Homework Statement

Write the vector expression for the acceleration a of the mass center G of the simple pendulum in both n-t and x-y coordinates for the instant when θ = 66° if θ'= 2.22 rad/sec and θ"= 4.475 rad/sec²

I have attached an image of the question.

Homework Equations

a_n = v²/r = rθ² = vθ'

a_t = v' = rθ'

The Attempt at a Solution

I've managed to calculate e_n and e_t correctly

a_t = (4.2 ft)(4.475 rad/sec²) = 18.795 ft/sec²

a_t = 18.795ft/sec²

For a_n I calculated velocity first:

v = rθ' = (4.2ft)(2.22 rad/sec)
v = 9.324 ft/sec

Hence a_n = (9.324 ft/sec)(2.22 rad/sec)
a_n = 20.69928 ft/sec²

Unfortunately, I'm now having difficulty with finding the velocity in terms of i and j.

I had thought that I could use geometry to do it:

a_tcos(90-66) = -17.77 i
a_tsin(90-66) = 7.644 j

But the system says it's wrong. Help is greatly appreciated.

Both n and t components contribute to each of the i and j components.

 

[Northbysouth]

@SammyS: You were right. To find the i component I did the following:

-a_tcos(24) -a_ncos(66) = -25.589i

For j:

a_tsin(24) - a_nsin(66) = -11.265 j

Thanks everyone.

 

[Nickie]

The vector expression for the acceleration a of the mass center G of the simple pendulum can be written as follows:

In n-t coordinates:

a = an n + at t

where an is the normal component of acceleration and at is the tangential component of acceleration.

In x-y coordinates:

a = ax i + ay j

where ax is the x-component of acceleration and ay is the y-component of acceleration.

Using the given values, we can calculate the components of acceleration in both coordinate systems:

In n-t coordinates:

an = vθ' = (4.2 ft)(2.22 rad/sec) = 9.324 ft/sec
at = v' = rθ" = (4.2 ft)(4.475 rad/sec2) = 18.795 ft/sec2

Therefore, the vector expression for acceleration in n-t coordinates is:

a = (9.324 ft/sec) n + (18.795 ft/sec2) t

In x-y coordinates:

ax = atcosθ - ansinθ = (18.795 ft/sec2)cos(66°) - (9.324 ft/sec)sin(66°) = 7.644 ft/sec2
ay = atsinθ + ancosθ = (18.795 ft/sec2)sin(66°) + (9.324 ft/sec)cos(66°) = 17.77 ft/sec2

Therefore, the vector expression for acceleration in x-y coordinates is:

a = (7.644 ft/sec2) i + (17.77 ft/sec2) j

I hope this helps! Keep up the good work in your studies of physics.