A bob of mass m = 0.250 kg is suspended from a fixed point with a massless strin

Question

A bob of mass m = 0.250 kg is suspended from a fixed point with a massless string of length = 25.0 cm. You will investigate the motion in which the string traces a conical surface with half-angle theta = 17.0 degrees.

What tangential speed must the bob have so that it moves in a horizontal circle with the string making an angle 17.0 degrees with the vertical? Express answer in meters per second.

A bob of mass m = 0.250 kg is suspended from a fixed point with a massless string of length = 25.0 cm. You will investigate the motion in which the string traces a conical surface with half-angle theta = 17.0 degrees. What tangential speed must the bob have so that it moves in a horizontal circle with the string making an angle 17.0 degrees with the vertical? Express answer in meters per second.

Explanation / Answer

The only force acting in the plane of the circular path is the horizontal component oftension in the string. It is Tsin (sine because the angle is measured from the vertical). The force causes a centripetal acceleration since the mass moves in a circular path. Using Newton’s 2nd law, we can write:

F(x)= mv²/r = Tsin---------->(1)

There are two forces acting on the mass vertically, the vertical (upward) component of tension (Tcos) and the weight (mg) of the mass acting downward:

F(y) = 0 = Tcos - mg
mg = Tcos-------------->(2)

Dividing (1) by (2) (note that mass and tension divides out), you obtain:

v = [grtan]

Of course we need to find the radius, and it is r = Lsin (you should be able to see why), so:

r = 0.250msin17.0°
= 0.0731m

Therefore, the required speed is

v = [(9.80m/s²)(0.0731m)tan17.0°]
= 0.468m/s

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