Question

Will somebody help explain to me and teach me: Rotational Motion and help explain how to use the formulas? (I need help on changing word problems into equations)

Answer 1

@AllTehMaffs

Answer 2

@Mertsj

Answer 3

@hartnn

Answer 4

@kittiwitti1

Answer 5

@wolfe8

Answer 6

Can you give an example of question. I cannot promise I will be able to do it though, but it will help us to figure out what to show you. That, or you can use this if it helps http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html

Answer 7

Find the centripetal accerlation (ac) of a roating wheel of radius 0.650m if it has
a. an angular velcoity of 5.00 rad/s

b. a tangetial velocity (vT) of 12 m/s

Answer 8

@wolfe8

Answer 9

Centripetal acceleration = radial distance * (angular speed)^2
where angular speed is in rad/s

Answer 10

Plug in the values given taking note of the units. Let me know what you get.

Answer 11

So if we look at a rotating object, (a disc!), there are two types of velocities that are associated with it.
For any point on the rotating object, the rotational velocity omega, ω is the measure of how many degrees or radians a point rotates through in a given amount of time.
\[ \omega = \frac{\Delta \theta}{\Delta t}\]
It has units of radians per second, or degrees per second (or some angular measurement per unit time).
\[\text{in this problem } \ \omega \propto rad/s\]
For any two points on the rotating body (in the picture, the smaller circle is just the path an object on the larger disc traces as it rotates 2pi radians, or 360º) the angular velocity is the same
\[ (\text{in the picture,} \ ω_1 = ω_2 )\]
- that is, it covers the same angular displacement (the "amount" of degrees, or radians) in the same amount of time.


The magnitude of the tangential velocity of a point on the rotating object is **larger the further away it is from the center of rotation
\[ (\text{in the picture,} \ v_2>v_1)\]
as can be seen from examination of the formula for tangential velocity
\[v_T = r\omega\]
which has units of distance per time (meters per second in the case of this problem).
\[\text{in this problem } \ v_T \propto m/s\]
The further away from the center the point is, the larger the r is, so the larger the value for tangential velocity becomes.
This can be seen as a conversion, given that there are 2 pi radians in one complete circumference of a circle
\[\left(\frac{rad}{s}\right) \left( \frac{2\pi r}{2 \pi rad}\right)\]

Because the angular velocity must remain constant, the tangential velocity must increase the further away from the point of rotation you measure, since it has more distance to cover in the same amount of time (the circumference of the circle it traces is larger)

So! As for the question, we have a wheel of radius r=.65m.


For part a, we know that centripetal acceleration is defined to be
\[a_{cent} = \frac{v^2}{r}\]
noting that acceleration has units of meters per second squared
\[ a \propto m/s^2\]
and from our formula for tangential velocity, we know that
\[v_T = r \omega\]
so
\[a_{cent} = \frac{(r\omega)^2}{r}
\\ \
\\a_{cent} =r\omega^2\]

That does it for part a.

For part b, you have the easy job of plugging in the tangential velocity and the radius into the formula for centripetal acceleration.