Results 1 to 4 of 4

Math Help - Radial Velocity

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    239

    Radial Velocity

    A particle moves in a plane with constant radial velocity  \dot{r} = 4 \ \frac{m}{s} . The angular velocity is constant and has magnitude  \dot{\theta} = 2 \ \frac{rad}{s} . When the particle is  3 \ m from the origin, find the magnitude of (a) the velocity and (b) the acceleration.

    So  \bold{v} = v_{r} \bold{\hat{r}} + v_{\theta} \bold{\hat{\theta}} where the first component is the radial velocity and the second component is the angular velocity.

    So we want units in  \frac{m}{s} (same units). So would I just change the  2 \ \frac{rad}{s} into  \frac{m}{s} and take the magnitude?

    If we have constant velocity (both radial and angular) wouldn't the magnitude of the acceleration be  0 ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2007
    Posts
    239
    I think  ||\bold{v}|| = \sqrt{36} \ \frac{\text{m}}{\text{s}} ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2007
    Posts
    239
    is this too difficult?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,843
    Thanks
    320
    Awards
    1
    Quote Originally Posted by shilz222 View Post
    A particle moves in a plane with constant radial velocity  \dot{r} = 4 \ \frac{m}{s} . The angular velocity is constant and has magnitude  \dot{\theta} = 2 \ \frac{rad}{s} . When the particle is  3 \ m from the origin, find the magnitude of (a) the velocity and (b) the acceleration.

    So  \bold{v} = v_{r} \bold{\hat{r}} + v_{\theta} \bold{\hat{\theta}} where the first component is the radial velocity and the second component is the angular velocity.

    So we want units in  \frac{m}{s} (same units). So would I just change the  2 \ \frac{rad}{s} into  \frac{m}{s} and take the magnitude?

    If we have constant velocity (both radial and angular) wouldn't the magnitude of the acceleration be  0 ?
    The velocity is the vector sum of the radial and angular components. That is to say
    \vec{v} = \dot{r} \bold{\hat{r}} + r \dot{\theta} \bold{\hat{\theta}}

    Take the time derivative of this and you get:
    \vec{a} =(\ddot{r} - r \dot{\theta}^2) \bold{\hat{r}} + (2\dot{r} \dot {\theta} + r \ddot{\theta})\bold{\hat{\theta}}

    Note the "r" next to the \theta in the velocity equation. This is where your units were inconsistent. The second term in the expansion for the acceleration is the centripetal acceleration term, which will be present even if there is no radial or angular accelerations. (The "complicated" form of the acceleration is due to the change in the unit vector \hat{\theta} over time.)

    -Dan
    Last edited by topsquark; November 29th 2007 at 04:00 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Radial heat flow in a cylinder
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: February 2nd 2011, 02:30 PM
  2. Radial Coordinate and Partial Differentiation
    Posted in the Calculus Forum
    Replies: 0
    Last Post: January 23rd 2010, 12:30 PM
  3. How to calculate length of a radial line?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 27th 2009, 08:36 AM
  4. radial force field...
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 24th 2008, 06:56 PM
  5. Curl of a Radial field
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 14th 2008, 07:14 AM

Search Tags


/mathhelpforum @mathhelpforum