- #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.
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?
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?