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Posted by on Wednesday, February 15, 2012 at 9:31pm.

A particle is moving clockwise in a circle of radius 1.54 m. At a certain instant, the magnitude of its acceleration is a = 26.0 m/s2,and the acceleration vector has an angle of θ = 50° with the position vector, as shown in the figure. At this instant, find the speed, v of this particle.

  • physics - , Wednesday, February 15, 2012 at 9:44pm

    a*sin50=v^2/r solve for v

  • physics - , Wednesday, February 15, 2012 at 9:48pm

    not the right answer

  • physics - , Thursday, February 16, 2012 at 7:49am

    It seems to me that a*cos50 is the centripetal component of the acceleration. The position vector is in the radial direction.

    a*cos50 = v^2/r = 16.71 m/s^2
    v^2 = 25.74 m^2/s^2
    v = 5.07 m/s

  • physics - , Thursday, February 16, 2012 at 12:14pm

    thanks

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