# Angle of acceleration in non-uniform circular motion

• member 731016
In summary, the conversation is about calculating the angle of a line using the inverse tangent function. The OP is unsure about the notation used and the symbol for degrees in the solution. It is clarified that the symbol used is a gamma, which represents degrees. The solution also involves finding the angle between a radial acceleration and total acceleration, which is measured anticlockwise from the cord.
member 731016
Homework Statement
Pls see below
Relevant Equations
Pls see below
For (c),

Solution is

Can someone please explain how they calculated that angle? I thought they would do ##arc\tan (\frac {32}{3.35})##

Many thanks!

Callumnc1 said:
View attachment 322504
Can someone please explain how they calculated that angle? I thought they would do ##arc\tan (\frac {32}{3.35})##

Many thanks!
##\tan^{-1}## is one way of writing ##\arctan##.
Or is it the 3.35/32 instead of the other way up that bothers you?

member 731016 and MatinSAR
haruspex said:
##\tan^{-1}## is one way of writing ##\arctan##.
Or is it the 3.35/32 instead of the other way up that bothers you?
The OP seems to be unaware that the trig function inverses can be written in two ways: e.g., ##\tan^{-1}(x)## is the same as arctan(x), and similar for the other circular trig functions.

On another note, does anyone recognize the symbol that follows 5.98 in the solution? Evidently it means "degrees" since ##\tan^{-1}(\frac{3.35}{32.0}) \approx 5.98## (degrees), but the symbol used looks like ##\gamma## to me, which I've never seen used to signify degrees .

member 731016 and MatinSAR
haruspex said:
##\tan^{-1}## is one way of writing ##\arctan##.
Or is it the 3.35/32 instead of the other way up that bothers you?

Sorry I should been more accurate - it is the 3.35/32 that is bothering me

Many thanks!

Mark44 said:
The OP seems to be unaware that the trig function inverses can be written in two ways: e.g., ##\tan^{-1}(x)## is the same as arctan(x), and similar for the other circular trig functions.

On another note, does anyone recognize the symbol that follows 5.98 in the solution? Evidently it means "degrees" since ##\tan^{-1}(\frac{3.35}{32.0}) \approx 5.98## (degrees), but the symbol used looks like ##\gamma## to me, which I've never seen used to signify degrees .

Sorry I was actually aware that ##tan^{-1} = arctan##. However, I was not sure why they wrote 3.35\32 instead 32\3.35 inside the tan function.

Also yeah that textbook solutions dose use the gamma symbol for degrees for some reason.

Many thanks!

Callumnc1 said:

Sorry I should been more accurate - it is the 3.35/32 that is bothering me

Many thanks!
The 32 acceleration is radial. The angle quoted is that between the radial acceleration and the total acceleration. The cosine of that angle would be radial/total and its tangent is tangential/radial.

member 731016
haruspex said:
The 32 acceleration is radial. The angle quoted is that between the radial acceleration and the total acceleration. The cosine of that angle would be radial/total and its tangent is tangential/radial.
Thank you for reply @haruspex !

When I draw a diagram of the accelerations acting on the mass when ##\theta = 20~degrees##

Then draw a vector addition diagram to get the total acceleration,

I see that ##\tan\theta = \frac{a_c}{a_t} = \frac{32}{3.35}##

I know this is bad practice to change symbols when solving a problem, but if I add an angle phi (use to represent tangent/radial acceleration) ,

Then I can see that ##\tan\phi = \frac{a_t}{a_c}##. I think I'm now not sure how they got acceleration to be below the cord at 5.89 degrees.

Many thanks!

Callumnc1 said:
View attachment 322528
Many thanks!
The angle of a line "below the cord" means the angle between the cord and the line, measured on the underside of the cord. In your last diagram above, that is the angle between the pale brown cord and the black total acceleration vector, measured anticlockwise from the cord.

member 731016
haruspex said:
The angle of a line "below the cord" means the angle between the cord and the line, measured on the underside of the cord. In your last diagram above, that is the angle between the pale brown cord and the black total acceleration vector, measured anticlockwise from the cord.
Oh thank you for your help @haruspex! I see it now :)

#### Attachments

• 1676846016881.png
5.9 KB · Views: 70

## What is the angle of acceleration in non-uniform circular motion?

The angle of acceleration in non-uniform circular motion refers to the angle between the tangential acceleration and the centripetal (radial) acceleration. These two components of acceleration are perpendicular to each other, so the angle is typically 90 degrees.

## How do you calculate the tangential acceleration in non-uniform circular motion?

Tangential acceleration ($$a_t$$) can be calculated using the formula $$a_t = \alpha \cdot r$$, where $$\alpha$$ is the angular acceleration and $$r$$ is the radius of the circular path. Tangential acceleration is responsible for the change in the magnitude of the velocity of the object.

## What role does centripetal acceleration play in non-uniform circular motion?

Centripetal acceleration ($$a_c$$) is responsible for changing the direction of the velocity vector, keeping the object moving in a circular path. It is given by the formula $$a_c = \frac{v^2}{r}$$, where $$v$$ is the instantaneous tangential velocity and $$r$$ is the radius of the circular path.

## How do you find the total acceleration in non-uniform circular motion?

The total acceleration ($$a$$) in non-uniform circular motion is the vector sum of the tangential acceleration ($$a_t$$) and the centripetal acceleration ($$a_c$$). Since these two components are perpendicular, the magnitude of the total acceleration can be found using the Pythagorean theorem: $$a = \sqrt{a_t^2 + a_c^2}$$.

## Can the angle of acceleration in non-uniform circular motion change?

No, the angle between tangential and centripetal acceleration remains constant at 90 degrees because these two components are always perpendicular to each other. However, the magnitudes of these accelerations can change, affecting the overall acceleration vector.

• Introductory Physics Homework Help
Replies
55
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
376
• Introductory Physics Homework Help
Replies
16
Views
456
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
32
Views
1K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
470