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Find the resultant acceleration of a particle moving on a circle of radius $0.70\ m$, if its angular speed is 37 rpm and its tangential acceleration is $2.9 \frac m{s^2}$.

Express the angle with respect to the tangential velocity vector.

I thought that I need to start out finding the circumference fist by $2\pi \cdot 0.7 = 1.4\cdot \pi$.

Then take $\frac {1.4\cdot\pi\cdot37}{ 60\,sec} = 2.7122$.

However, this is wrong and I have no idea how to figure the tangential velocity vector. Can anyone help me out on this?

Community
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Shane Yost
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3 Answers3

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The acceleration has two perpendicular components. The tangential acceleration is given as $a_t=2.9m/s^2$. The centripetal acceleration (that keeps a body moving in a circle) is given by $a_c=\omega^2r$. Here $\omega$ is the angular velocity in radians per second. Every rotation has an angle of $2\pi$, so $\omega=2\pi\cdot 37/60\ rad/s$. The angle between the acceleration and the tangential vector is $\theta$. If you draw the two components you will see that $$\tan\theta=\frac{a_c}{a_t}$$ Calculate first $\omega$, then $a_c$, and finally plug them into the above equation.

Andrei
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We can use the following parametric equation to describe the motion of the object: $$x=r \big(\cos(\omega t),\sin(\omega t)\big)$$ Where $r$ is the radius of the circle, and $\omega$ is the angular velocity of the object.
From this equation, we can get the following:$$\dot{x}=\dot r \big(\cos(\omega t),\sin(\omega t)\big)+r\omega \big(-\sin(\omega t),\cos(\omega t)\big)$$ Since $r$ is constant, we conclude that $\dot{r}=0$. So, we have the following: $$\dot{x}=r\omega \big(-\sin(\omega t),\cos(\omega t)\big)$$ Differentiating again with respect to $t$, we get the following: $$\ddot{x}=\left(r\dot{w}+\dot{r}w\right)\big(-\sin(\omega t),\cos(\omega t)\big)-r\omega^2\big(\cos(\omega t),\sin(\omega t)\big)$$ Again, since $\dot{r}=0$, we can simplify: $$\ddot{x}=r\dot{w}\big(-\sin(\omega t),\cos(\omega t)\big)-r\omega^2\big(\cos(\omega t),\sin(\omega t)\big)$$ We notice that $\ddot{x}$ is the resultant acceleration. The first term on the right is the tangential acceleration, while the second is the centripetal acceleration. Notice that they are perpendicular to each other. You can use the Pythagorean Theorem to find the magnitude of the resultant acceleration. This only works if $\omega$ is measured in radians per second.

Hrhm
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First, you need to find the centripetal acceleration of the particle. This is a function of the radius in meters and the speed in meters per second, which you've already computed. Do you know the formula that expresses centripetal acceleration in these terms?

Then, the resultant acceleration is a vector sum of the centripetal acceleration and the tangential acceleration. Can you draw the relevant diagram?

(Edit: fixed the units of speed, since you have meters per second.)

Chris Culter
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  • I'm guessing the centripetal acceleration formula is < a = (rw^2)/r = rw^2 >. so a = .7(2.7122)^2 + 2.9 – Shane Yost Sep 25 '13 at 23:13
  • @ShaneYost Not quite. The Greek letter that looks like a "w" is omega, $\omega$, and it represents angular speed in radians per second, or units of 1/s. Your quantity $v = 2.7122$ is actually a speed in meters per second, or units of m/s. You can catch this kind of mistake by always preserving units in your intermediate calculations. Anyway, the relationship between the two is $v=r\omega$. If you want to use $v$ directly, the formula is $a=v^2/r$. – Chris Culter Sep 25 '13 at 23:21