Circular motion with kinematics

jisbon

Homework Statement
Consider object going through circular motion with radial acceleration = $2m/s$ and radius given by $\frac{4}{2t+2}$ Find arc length of object swept through the first two seconds.
Homework Equations
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My working:

$s=\int v$
$v= \sqrt{\frac{a_{c}}{r}}=\sqrt{\frac{a_{c}}{\frac{4}{2t+2}}}$
$s= \int_{0}^{2} \sqrt{\frac{2}{\frac{4}{2t+2}}}$
My final answer seems to be wrong. Any ideas? Cheers

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Delta2

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I find the statement contradictory. If it is circular motion then the radius must be constant in time (radial acceleration might not be constant).

haruspex

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The radial acceleration (radial component of total acceleration) has two subcomponents. $r\dot\theta^2$ is one; what is the other?
Are you sure you have quoted r=r(t) correctly? There's a factor of 2 common to numerator and denominator.

jbriggs444

Homework Helper
Homework Statement: Consider object going through circular motion with radial acceleration = $2m/s$
The units of measurement for that acceleration do not appear to be correct. Also, if the motion is circular, how can the radius be changing over time?

Orodruin

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The units of measurement for that acceleration do not appear to be correct. Also, if the motion is circular, how can the radius be changing over time?
That the acceleration in the radial direction is non-zero does not mean that the radius is changing. On the contrary, if there was no acceleration in the radial direction then the motion would be linear, not circular.

jbriggs444

Homework Helper
That the acceleration in the radial direction is non-zero does not mean that the radius is changing. On the contrary, if there was no acceleration in the radial direction then the motion would be linear, not circular.
The original post quotes a formula for radius that depends on t. Such is incompatible with circular motion.

Edit: Unless one is describing circular motion about one point using polar coordinates about an origin elsewhere? In which case, the terms "radius" and "radial" are hideously misleading.

Last edited:

kuruman

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Perhaps "circular" is a misnomer and "curvilinear" is more appropriate to resolve the issue of the time-varying radius.
$v= \sqrt{\frac{a_{c}}{r}}=\sqrt{\frac{a_{c}}{\frac{4}{2t+2}}}$
Your expression for $v$ in terms of $a_c$ is incorrect.

My final answer seems to be wrong. Any ideas? Cheers
You don't show a final answer. What is it? Also, please get into the habit of writing integrals with a d(something) included.

jisbon

The radial acceleration (radial component of total acceleration) has two subcomponents. $r\dot\theta^2$ is one; what is the other?
Are you sure you have quoted r=r(t) correctly? There's a factor of 2 common to numerator and denominator.
Yes, it was quoted correctly. $r(t) = \frac {4}{2t+2} jisbon Perhaps "circular" is a misnomer and "curvilinear" is more appropriate to resolve the issue of the time-varying radius. Your expression for$v$in terms of$a_c$is incorrect. You don't show a final answer. What is it? Also, please get into the habit of writing integrals with a d(something) included. Replying to your 1st statement: Realized its:$v_{tan} =\sqrt{ra_{c}}\int_{0}^{2} \sqrt{ra_{c}} \int_{0}^{2} \sqrt{2*\frac{4}{2t+2}} dt$= 4.39444.... Which is still incorrect unfortunately. The answer seems to be 2.93 TSny Homework Helper Gold Member$\int_{0}^{2} \sqrt{2*\frac{4}{2t+2}} dt$= 4.39444.... Which is still incorrect unfortunately. The answer seems to be 2.93 Check the evaluation of this integral. I think it yields 2.93 (Did you neglect the square root when evaluating the integral?) Note that you assumed that the centripetal acceleration is 2 m/s2 and that the radius of curvature is given by$\frac{4}{2t+2}##. Unfortunately, the wording of the problem statement does not make this interpretation clear. "radial acceleration" is generally not the same as centriptetal acceleration, and "radius" does not necessarily indicate radius of curvature. Also, the motion is clearly not circular motion.

• Delta2

"Circular motion with kinematics"

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