What Is Gaetan Boucher's Total Acceleration at 1/4 Lap?

[Lolagoeslala]

Homework Statement

Gaetan Boucher started from rest and skated around a circular ice track of radius 100m. He maintained a constant rate of increase of his speed and finished one complete lap in 30s. Calculate the magnitude of his total acceleration as he passed a point 1/4 of the distance around the track from the start.

The Attempt at a Solution

v1 = 0m/s
r = 100m
d = 1/4d
a = ?
t = 30s

c = 1/4 x 2rπ
c = 157 m

t = 1/4 (30s)
t = 7.5s
157 m = (0m/s + v2)/2 (7.5s)
41.866 m/s = v2

157 m = 1/2a(t)2
5.58 m/s^2 = a

 

[rude man]

There are two accelerations to be solved for here: tangential and radial.

First get tangential accel: this should be easy from the fact that this accel is constant.

Then you need to solve for speed v at the 1/4 point. You can do that by solving for the time it takes him to go to the 1/4 point (again using the tangential accel.) , then use your value of tangential accel one more time to get v. Finally, you compute the other component of accel (hint: it's called centripetal) to get the total accel.

The total accel is not the numerical sum of the two. The two accels are separate vectors, one pointing tangential to his course and the other towards the center of the ring. If you don't know vectors just leave the answer as two separate accelerations.

 

[Lolagoeslala]

There are two accelerations to be solved for here: tangential and radial.

First get tangential accel: this should be easy from the fact that this accel is constant.

Then you need to solve for speed v at the 1/4 point. You can do that by solving for the time it takes him to go to the 1/4 point (again using the tangential accel.) , then use your value of tangential accel one more time to get v. Finally, you compute the other component of accel (hint: it's called centripetal) to get the total accel.

The total accel is not the numerical sum of the two. The two accels are separate vectors, one pointing tangential to his course and the other towards the center of the ring. If you don't know vectors just leave the answer as two separate accelerations.

Okay i am a bit confused.
you said tangential acceleration,
a = (rw) / t

right?
i have not learned the tangential acceleration thoroughly so.

 

[SammyS]

Okay i am a bit confused.
you said tangential acceleration,
a = (rw) / t

right?
i have not learned the tangential acceleration thoroughly so.

What is the angular acceleration ?

 

[Lolagoeslala]

What is the angular acceleration ?

using this formula ω = θ / t

but i do not know the angle .. not wait i do.
1/4 x 360 = 90 deg

ω = θ / t
ω = n/2 / 7.5s

 

[SammyS]

using this formula ω = θ / t

but i do not know the angle .. not wait i do.
1/4 x 360 = 90 deg

ω = θ / t
ω = n/2 / 7.5s

That's the average angular velocity for the first 1/4 of a lap.

The skater starts with ω_i = 0, & averages π/(2(7.5)) rad/s ?

What is ω_f at 1/4 lap ?

(Just as in the case of constant linear velocity, for constant angular velocity, ω_Average = (ω_i + ω_f)/2 .)

 

[Lolagoeslala]

That's the average angular velocity for the first 1/4 of a lap.

The skater starts with ω_i = 0, & averages π/(2(7.5)) rad/s ?

What is ω_f at 1/4 lap ?

(Just as in the case of constant linear velocity, for constant angular velocity, ω_Average = (ω_i + ω_f)/2 .)

π/(2(7.5)) rad/s = (0 +ωf / 2)
π/7.5 rad/s = ωf

 

[SammyS]

π/(2(7.5)) rad/s = (0 +ωf / 2)
π/7.5 rad/s = ωf

Yes.

From ω_f & ω_i , you should be able to find the angular acceleration, α .

 

[Lolagoeslala]

Yes.

From ω_f & ω_i , you should be able to find the angular acceleration, α .

π/7.5 rad/s = at
π = a

 

[SammyS]

π/7.5 rad/s = at
π = a

No. Bad algebra.

t = 7.5 s ,

so a = (π/7.5)/7.5 rad/s²

= ?

 

[Lolagoeslala]

No. Bad algebra.

t = 7.5 s ,

so a = (π/7.5)/7.5 rad/s²

= ?

a= π/56.25 rad/s^2

 

[SammyS]

a= π/56.25 rad/s^2

The tangential component of acceleration is \ \ a_t=\alpha R\ .

The radial component of acceleration is simply the centripetal acceleration.

 

[Lolagoeslala]

The tangential component of acceleration is \ \ a_t=\alpha R\ .

The radial component of acceleration is simply the centripetal acceleration.

like this?

π/56.25 rad/s^2 x 100 m = a
5.58 radm/s^2 = a

 

[SammyS]

like this?

π/56.25 rad/s^2 x 100 m = a
5.58 radm/s^2 = a

That's only the tangential component.

 

[Lolagoeslala]

That's only the tangential component.

a = 5.58 radm/m^s2
v at the 1/4 point. You can do that by solving for the time it takes him to go to the 1/4 point
d = 200 m
C = 200 m x 3.14
C = 628 m
D = 1/4 628 m = 157 m

157 m = 1/2 (5.58 radm/m^s2)(t)^2
t = 7.5 s

157 m x 2 / 7.5 s =v2
41.866 m/s = v2

a = 41.866 m/s /100m
a = 0.041866 m/s^2

 

[SammyS]

a = 5.58 radm/m^s2
v at the 1/4 point. You can do that by solving for the time it takes him to go to the 1/4 point
d = 200 m
C = 200 m x 3.14
C = 628 m
D = 1/4 628 m = 157 m

157 m = 1/2 (5.58 radm/m^s2)(t)^2
t = 7.5 s

157 m x 2 / 7.5 s =v2
41.866 m/s = v2

a = 41.866 m/s /100m
a = 0.041866 m/s^2

You went through some needless steps there.

You already have ω at t = 7.5s, which you already know is the time at the 1/4 lap mark.

v = ωR .

Centripetal acceleration is a_c = v²/R .

(It's also equal to ²R)

I get a value different from yours.

By the way, rad∙m/s² is merely m/s² .

 

[Lolagoeslala]

You went through some needless steps there.

You already have ω at t = 7.5s, which you already know is the time at the 1/4 lap mark.

v = ωR .

Centripetal acceleration is a_c = v²/R .

(It's also equal to ²R)

I get a value different from yours.

By the way, rad∙m/s² is merely m/s² .

157 m x 2 / 7.5 s =v2
41.866 m/s = v2
a = v^2 / R
a = 41.866 m/s^2 / 100m
a = 17.52 m/s^2

 

[SammyS]

157 m x 2 / 7.5 s =v2
41.866 m/s = v2

a = v^2 / R
a = 41.866 m/s^2 / 100m
a = 17.52 m/s^2

That's about right.

Tangential acceleration is in the direction of motion. Centripetal acceleration is towards the center of the circle. Therefore, they're perpendicular to each other.

Find the magnitude of the acceleration using your two results.

 

[Lolagoeslala]

a = 17.52 m/s^2[W] + 5.58 m/s^2
like this?

 

[SammyS]

a = 17.52 m/s^2[W] + 5.58 m/s^2
like this?

No.

Acceleration is a vector quantity.

Tangential component is perpendicular to radial.

 

[Lolagoeslala]

No.

Acceleration is a vector quantity.

Tangential component is perpendicular to radial.

i am bit confused i have not worked with tangential component before

 

[SammyS]

i am bit confused i have not worked with tangential component before

In earlier posts this was mentioned, particularly in post #12.

Consider an object moving along some path. The velocity vector for the object will be along a line tangent to the path. The component of acceleration in the direction of the velocity vector is given by the time rate of change of the speed of the object: \displaystyle \frac{dv}{dt}\,,\ \ \text{ where }\ v=|\vec{v}|\ .

At any rate, the tangential component is perpendicular to radial, so use the Pythagorean theorem to find the magnitude.

 

[Lolagoeslala]

c^2 = (17.52 m/s^2) ^2 + (5.58 m/s^2 )^2
c = 18.38 m/s^2

 

[SammyS]

c^2 = (17.52 m/s^2) ^2 + (5.58 m/s^2 )^2
c = 18.38 m/s^2

That's correct to 3 sig. fig.

 

[Lolagoeslala]

That's correct to 3 sig. fig.

so that's the answer
i do not need to worry about the orientation?

 

[Lolagoeslala]

but the answer does not match up with the answers possible.

 

[SammyS]

but the answer does not match up with the answers possible.

What are the possible answers ?

 

[Lolagoeslala]

What are the possible answers ?

1.4 , 2.99 , 4.39 , 4.60 , and 5.78

 

[SammyS]

Oh ! I see what went wrong.

It takes longer than 7.5 s for the first 1/4 of a lap.

Base the angular acceleration on one complete lap.

ωf = 2ω_avg = 2(2π)/30 .

Angular acceleration, α = (ω_f-0)/30 . That's 1/4 of the former answer.

 

[Lolagoeslala]

Oh ! I see what went wrong.

It takes longer than 7.5 s for the first 1/4 of a lap.

Base the angular acceleration on one complete lap.

ωf = 2ω_avg = 2(2π)/30 .

Angular acceleration, α = (ω_f-0)/30 . That's 1/4 of the former answer.

i am abit confused

 

[SammyS]

i am abit confused

Confused about what in particular ?

 

[Lolagoeslala]

Confused about what in particular ?

ωf = 2ωavg = 2(2π)/30 .

Angular acceleration, α = (ωf-0)/30 . That's 1/4 of the former answer.

 

[SammyS]

ω_f = 2ω_avg = 2(2π)/30 .

Angular acceleration, α = (ω_f-0)/30 . That's 1/4 of the former answer.

What you didn't include from that post was:

Base the angular acceleration on one complete lap.​

All I did in the above QUOTE'ed text was to repeat what we had previously done, but used a complete lap since for a complete lap, we know the elapsed time. Once you have this value, you will need to find the velocity at the 1/4 lap point so that you can recalculate the radial component of acceleration, since that is also wrong.

Do you know the kinematic equations for constant angular acceleration?




If you work everything out, you would find that using our previous answer for angular acceleration, π/7.5/7.5, one lap would take only 15 seconds.

 

[Lolagoeslala]

What you didn't include from that post was:

Base the angular acceleration on one complete lap.​

All I did in the above QUOTE'ed text was to repeat what we had previously done, but used a complete lap since for a complete lap, we know the elapsed time. Once you have this value, you will need to find the velocity at the 1/4 lap point so that you can recalculate the radial component of acceleration, since that is also wrong.

Do you know the kinematic equations for constant angular acceleration?




If you work everything out, you would find that using our previous answer for angular acceleration, π/7.5/7.5, one lap would take only 15 seconds.

i do not know how to find the
constant angular acceleration

 

[Lolagoeslala]

i am getting 4.60

 

[SammyS]

i am getting 4.60

Yes.

Notice, that's 2/4 of your previous answer.