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Motion of particle suspended by string, in horizontal plane

(Please see the attached file)

A particle is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point O. The particle is set into motion, so that it describes a horizontal circle whose centre is vertically below O. The angle between the string and the vertical is theta, as shown in the diagram (see attachment).

(a) The particle completes 40 revolutions every minute. Show that the angular speed of the particle is 4.19 radians per second, to three significant figures.

(b) The radius of the circle is 0.2 metres. Find the magnitude of the acceleration of the particle.

(c) The mass of the particle is m kg and the tension in the string is T newtons.

(i) Draw a diagram showing the forces acting on the particle.

(ii) Explain why T cos theta = mg.

(iii) Find the value of theta, giving your answer to the nearest degree.


Solution Preview

Please refer to the attachment.

a) Angular speed ω = 2Πf where f is the frequency in revolutions per sec.

In this case, f = 40/60 rps. Hence, ω = 2Πx40/60 = 4.19 rad/sec

b) Centripetal acceleration = v2/R = ω2R = 4.192 x 0.2 = 3.51 m/s2

c) i) Forces acting on the particle ...

Solution Summary

The motion of particles suspended by strings in a horizontal plane is determined. Step by step solution provided.