Did I Calculate Velocity and Acceleration Correctly?

In summary, the conversation discusses finding the magnitude of velocity and acceleration at a given angle in a rotational motion problem. The process of calculating tangential velocity and acceleration is correct, but the centripetal velocity should be calculated using the equation v_{centripetal} = \omega * r where \omega is the angular velocity. The centripetal acceleration can then be calculated using the equation a_{centripetal} = \frac{v_{centripetal}^2}{r}.
  • #1
Lancelot59
646
1
This problem kicked me and I ended up using all my attempts. I just realized I probably failed this part because I used radian input while my calculator was set to degree mode.

Probs.12-175_176.jpg

OA rotated counterclockwise:
[tex]\theta'=3t^{\frac{3}{2}} rad/s[/tex]
When t=0 theta=0. Find the magnitude of velocity and acceleration when theta=30 degrees.

I went about finding some stuff first:
[tex]\theta=\frac{6}{5}t^{\frac{5}{2}} rad/s[/tex]
[tex]\omega=\theta'=3t^{\frac{3}{2}} rad/s[/tex]
[tex]\alpha=\theta''=\frac{9}{2}t^{\frac{1}{2}} rad/s[/tex]
And I isolated the applicable part of the curve:
[tex]r=+\sqrt{4cos(2\theta)[/tex]

I also solved for time at 30 degrees and got 0.717674 seconds.

I didn't fall into the trap of just using the angular velocity as the answer. I actually only could find the tangential velocity via this equation:
[tex]v_{tangental}=\omega*r[/tex]
[tex]v_{tangental}=3t^{\frac{3}{2}}*\sqrt{4cos(2\theta)[/tex]
[tex]v_{tangental}=3(0.717674)^{\frac{3}{2}}*\sqrt{4cos(2(\frac{\pi}{6}))[/tex]
[tex]v_{tangental}=2.87944975 m/s[/tex]
Which I got using the correct calculator mode this time.

PROBLEM:
How can I go about getting centripetal velocity?

As for acceleration I used this setup:
[tex]a_{tangental}=\alpha*r[/tex]
[tex]a_{tangental}=\sqrt{4cos(2\theta)*\frac{9}{2}*t^{\frac{1}{2}}[/tex]

[tex]a_{centripetal}=\frac{v_{tangental}}{r}[/tex]
[tex]a_{centripetal}=\frac{v_{tangental}}{\sqrt{4cos(2 \theta)}}[/tex]
Substituted in tangental velocity, took the magnitude and got a wrong answer, hopefully only because I had the wrong mode on my calculator.

Did I do this right or am I just silly?
 
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  • #2
The process you used to calculate the tangential velocity and acceleration is correct. However, when calculating the magnitude of the centripetal velocity, you should have used the equation:v_{centripetal} = \omega * r where \omega is the angular velocity. This is because the centripetal velocity is simply the product of the angular velocity and the radius (or in this case, the length of the arc). With that said, once you have calculated the centripetal velocity, you can calculate the centripetal acceleration by using the equation: a_{centripetal} = \frac{v_{centripetal}^2}{r}Hope this helps!
 

FAQ: Did I Calculate Velocity and Acceleration Correctly?

What is the difference between velocity and acceleration?

Velocity refers to the rate at which an object changes its position in a specific direction, while acceleration refers to the rate at which an object changes its velocity. In simpler terms, velocity is the speed and direction of an object, while acceleration is the change in speed or direction over time.

How do you calculate velocity and acceleration?

Velocity is calculated by dividing the change in position by the change in time, while acceleration is calculated by dividing the change in velocity by the change in time. These calculations can be represented mathematically as v = Δx/Δt for velocity and a = Δv/Δt for acceleration.

3. Can velocity and acceleration be negative?

Yes, both velocity and acceleration can be negative. This indicates that the object is moving in the opposite direction of the positive direction used in the calculation. For example, a car moving west would have a negative velocity if east is chosen as the positive direction.

4. What is the relationship between velocity and acceleration?

The relationship between velocity and acceleration is that acceleration is the rate of change of velocity. This means that a change in velocity will result in acceleration, and a constant velocity will result in zero acceleration. In other words, acceleration is the cause of changes in velocity.

5. How is velocity and acceleration represented on a graph?

Velocity is represented on a graph as the slope of the line, with the slope increasing or decreasing depending on the direction of the velocity. Acceleration is represented as the slope of the velocity line, with a steeper slope indicating a higher acceleration. Both velocity and acceleration can also be represented by vectors, with direction and magnitude indicated by arrows.

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