• Support PF! Buy your school textbooks, materials and every day products Here!

Finding the total acceleration

  • #1
217
0

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
 

Answers and Replies

  • #2
rude man
Homework Helper
Insights Author
Gold Member
7,766
760
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.
 
  • #3
217
0
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.
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
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 ?
 
  • #5
217
0
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
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
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 ωi = 0, & averages π/(2(7.5)) rad/s ?

What is ωf at 1/4 lap ?

(Just as in the case of constant linear velocity, for constant angular velocity, ωAverage = (ωi + ωf)/2 .)
 
  • #7
217
0
That's the average angular velocity for the first 1/4 of a lap.

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

What is ωf at 1/4 lap ?

(Just as in the case of constant linear velocity, for constant angular velocity, ωAverage = (ωi + ωf)/2 .)
π/(2(7.5)) rad/s = (0 +ωf / 2)
π/7.5 rad/s = ωf
 
  • #8
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
π/(2(7.5)) rad/s = (0 +ωf / 2)
π/7.5 rad/s = ωf
Yes.

From ωf & ωi , you should be able to find the angular acceleration, α .
 
  • #9
217
0
Yes.

From ωf & ωi , you should be able to find the angular acceleration, α .
π/7.5 rad/s = at
π = a
 
  • #10
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
π/7.5 rad/s = at
π = a
No. Bad algebra.

t = 7.5 s ,

so a = (π/7.5)/7.5 rad/s2

= ?
 
  • #11
217
0
No. Bad algebra.

t = 7.5 s ,

so a = (π/7.5)/7.5 rad/s2

= ?
a= π/56.25 rad/s^2
 
  • #12
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
a= π/56.25 rad/s^2
The tangential component of acceleration is [itex]\ \ a_t=\alpha R\ .[/itex]

The radial component of acceleration is simply the centripetal acceleration.
 
  • #13
217
0
The tangential component of acceleration is [itex]\ \ a_t=\alpha R\ .[/itex]

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
 
  • #14
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
like this?

π/56.25 rad/s^2 x 100 m = a
5.58 radm/s^2 = a
That's only the tangential component.
 
  • #15
217
0
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
 
  • #16
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
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 ac = 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 .
 
  • #17
217
0
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 ac = 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
 
  • #18
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
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.
 
  • #19
217
0
a = 17.52 m/s^2[W] + 5.58 m/s^2
like this?
 
  • #20
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
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.
 
  • #21
217
0
No.

Acceleration is a vector quantity.

Tangential component is perpendicular to radial.
i am bit confused i have not worked with tangential component before
 
  • #22
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
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: [itex]\displaystyle \frac{dv}{dt}\,,\ \ \text{ where }\ v=|\vec{v}|\ .[/itex]

At any rate, the tangential component is perpendicular to radial, so use the Pythagorean theorem to find the magnitude.
 
  • #23
217
0
c^2 = (17.52 m/s^2) ^2 + (5.58 m/s^2 )^2
c = 18.38 m/s^2
 
  • #24
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
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.
 
  • #25
217
0
That's correct to 3 sig. fig.
so thats the answer
i do not need to worry about the orientation?
 

Related Threads on Finding the total acceleration

Replies
1
Views
1K
Replies
2
Views
5K
Replies
2
Views
1K
Replies
2
Views
6K
Replies
3
Views
1K
Replies
3
Views
1K
Replies
2
Views
8K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
10K
  • Last Post
Replies
7
Views
3K
Top