A rock of mass m = 2.5 kg is tied to the end of a string of length L = 1.2m. The

Question

A rock of mass m = 2.5 kg is tied to the end of a string of length L = 1.2m. The other end of the string is anchored to the ceiling, and the rock is set in motion so that it travels in a horizontal circle of radius 0.70 m.
A) List all the forces acting on the rock and draw them onto a free body diagram of the rock.
B) Find the angle at which the rock is suspended. Also separate the tension on the string into vertical and horizontal components.
C) Apply Newtons second law in both the vertical and horizontal directions to find the horizontal acceleration? In which direction is this acceleration pointed?
D) Find the speed of the rock.

Explanation / Answer

A)I can't provide you the diagram, sorry for that
Forces acting on the rock:
1)Tension in the string
2)It's weight acting vertically downwards
3)Centripetal force
B) Let the angle be

cos=0.7/1.2

=54.3 deg

Vertical component of Tension=weight of the rock

=2.5(9.81)

=24.525N

Let Horizontal component of Tension be x

tan54.3=24.525/x

x=17.62N

C)F=ma

17.62=(2.5)a

a=7.05 m/s2

The acceleration is pointed towards the centre of the circular motion of the rock

D)a=v2/r

v2=ar

v2=(7.05)(0.7)

v2=.935

v=2.22 m/s

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