Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Circular motion and linear speed of an object

  1. Dec 3, 2006 #1
    An object is traveling around a circle with the radius of 5cm. If in 20sec the central angle of 1/3 radian is swept out, what is the angular speed of the object? Linear speed?

    Here's how I did it. angular speed-->
    a) omega=theta(in radians)/elapsed time
    = π/3/20= π/3*1/20 = π/60 radians/sec. Is that the same as the answer in the book, 1/60?

    linear speed-->
    b) v=rw (length/radius)(omega=angular speed)
    =5cm*(1/60)= 1/12 cm/s
    the answer in the book is 12m/s.

    what did I do wrong?
     
  2. jcsd
  3. Dec 3, 2006 #2
    Where did you get the n from? You went 1/3 radian in 20 seconds. To get the radian speed just find the distance traveled in one second. The book is right.

    If it really was 12 m/s it would have gone around the circle about 60 times in 20 seconds since the length of the circumference is 2 * pi * radius or .314 m. So the book is wrong. You are wrong too. Remember a radian is radius / arch distance. So r/s = 1/3 in 20 seconds. Once you have arch distance traveled in 20 seconds, finding linear speed is easy.
     
  4. Dec 4, 2006 #3

    HallsofIvy

    User Avatar
    Science Advisor

    interested learner, I don't think it was "n", it was [itex]\pi[/itex] using an overly simple font.

    ragbash, your problem says "1/3 radian". For some reason you used "[itex]\pi/3[/itex] radians.

    No, [itex]\pi/3[/itex] is not the same as 1/3!

    The circumference of a circle is [itex]2\pi r[/itex] or, since r= 5 here, [itex]10\pi[/itex] cm. Since the object moves 1/60 radian/sec and there are [itex]2\pi[/itex]radians in a circle, it is moving at [itex]\frac{1}{120\pi}[/itex] "circles per second" and so [itex]\frac{1}{120\pi}(10\pi)= 1/12[itex] cm/sec.
    Your book apparently has a typo for the second.
     
    Last edited by a moderator: Dec 4, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook