Tangential Acceleration/Tension

  • Thread starter srekai
  • Start date
In summary, the conversation was about a homework problem involving a car with a fuzzy die hanging from the rear view mirror, a circular curve with radius R, and an angle θ formed between the die and the vertical. The student attempted to solve the problem using the equations for force and summed forces, but made an error in solving for tan θ. Another user pointed out the mistake and suggested an alternative method using a right triangle formed by the tension force, gravity, and tangential force. The original student had some questions about the tangential force and its direction, to which the other user clarified that it should be referred to as a centrifugal force in the non-inertial frame of the car.
  • #1
srekai
8
0
I did the problem, just want to validate my solution/approach

1. Homework Statement

Your car happens to have a fuzzy die hanging from the rear view mirror. As you round a curve that is approximately a circle of radius R you notice that the fuzzy die makes an angle θ with the vertical. What is the angle θ as a function of R and your speed v?

Homework Equations


$$\sum F = ma$$
$$F = \frac{mv^2}{r}$$

The Attempt at a Solution


There are 3 forces on the dice as shown in the free body diagram attachment
FBD.png


From this the sum of the vertical forces and horizontal forces must cancel out, so we can say
$$F_{\text{tension,y}} = F_{\text{gravity}}$$
$$F_{\text{tension}} \cdot cos \theta = F_{\text{gravity}}$$
$$F_{\text{tension}} \cdot cos \theta =mg$$
$$F_{\text{tension}}= \frac{mg}{cos \theta}$$
and
$$F_{\text{tension,x}} = F_{\text{tangential}}$$
$$F_{\text{tension}} \cdot sin \theta = F_{\text{tangential}}$$
$$F_{\text{tension}} \cdot sin \theta = \frac{mv^2}{r}$$
$$ \frac{mv^2}{r} = mg \cdot \frac{sin \theta}{cos \theta}$$
$$tan \theta = \frac{v^2}{rg}$$
$$\theta = tan^{-1}\frac{v^2}{rg}$$
 

Attachments

  • FBD.png
    FBD.png
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Last edited:
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  • #2
Your idea is fine but you made an error in solving for ##\tan(\theta)## (check the units, you cannot take the arctan of a dimensionful number!). Alternatively you can immediately conclude that the tension force, gravity, and tangential force must form a right triangle where the tension force is the hypothenuse, the gravitational force is the side closest to the angle ##\theta##, and the tangential force the far side. It directly follows that
$$
\tan(\theta) = \frac{F_{\rm tang}}{F_g} = \ldots
$$
(I am leaving the dots for you to get the correct expression yourself)

Edit: Note that ##gv^2/r## has dimension ##\mathsf{L^2/T^4}##, i.e., not dimensionless.
 
  • #3
srekai: Your solution is correct. v^2 / rg is in fact dimensionless.
 
  • #4
Dr Dr news said:
srekai: Your solution is correct. v^2 / rg is in fact dimensionless.
He obviously edited that after my post.
 
  • #5
I have one remark and one question.
1. The component F sinθ should be in the radial direction, not tangential. It is pointing towards the center of the circle.
2. What is exerting the tangential force? I see only two forces, the tension and gravity.
 
  • #6
Chandra Prayaga said:
I have one remark and one question.
1. The component F sinθ should be in the radial direction, not tangential. It is pointing towards the center of the circle.
2. What is exerting the tangential force? I see only two forces, the tension and gravity.
All that is wrong in @srekai's (edited) solution is the reference to "tangential" force. A tangential force would be normal to the string, and as you say there is no such force present. But if we label it centrifugal force instead it all works.
 
  • #7
Absolutely. Agreed, provided the student is aware that the centrifugal force is present only in the noninertial frame of the car. In an inertial frame, the student's algebra would still be fine with only the tension and gravity present, and Newton's second law is applied
 

Related to Tangential Acceleration/Tension

What is tangential acceleration?

Tangential acceleration is the rate of change of an object's tangential velocity, or the velocity in the direction of motion along a curved path.

How is tangential acceleration calculated?

Tangential acceleration can be calculated by dividing the change in tangential velocity by the change in time.

What is the relationship between tangential acceleration and centripetal acceleration?

Tangential acceleration and centripetal acceleration are both components of an object's total acceleration along a curved path. Tangential acceleration is in the direction of motion, while centripetal acceleration is perpendicular to the direction of motion, towards the center of the curve.

What is tension in relation to tangential acceleration?

Tension is the force exerted by a string or rope on an object that is attached to it. In the context of tangential acceleration, tension is the force that causes a change in the object's tangential velocity.

How does tangential acceleration affect an object's motion?

Tangential acceleration can cause an object to speed up, slow down, or change direction along a curved path. It is an important factor in circular motion and can also be present in other types of motion, such as when an object is moving along a curved surface.

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