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The acceleration of an object going at a constant speed around a circle is

a = v^2 / r ,

and the acceleration is directed toward the center of the circle

Therefore:

• a = (30 cm/s)^2 / 20cm = 45 cm/s^2

Net force = mass * acceleration, and the centripetal force is the net force.  Thus

• centripetal force = mass * centripetal acceleration.

The object's mass is 110g = .110kg, and 45 cm/s^2 = .45 m/s^2 so

• F_cent = m * a_Cent = .110kg * .45 m/s^2 = .0495 kg * m/ s^2 = .0495 N

Since net force = mass * acceleration, an acceleration toward the center implies a net force toward the center.

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A simple pendulum has length 2 meters.  It is pulled back 10 cm from its equilibrium position and released.  The tension in the string is 5 Newtons.

Sketch the system with the pendulum mass at the origin and the x axis horizontal.

Sketch a vector representing the direction of the pendulum string at this instant.  As measured from a horizontal x axis, what is the direction of this vector?  (Hint:  The y component of this vector is practically the same as the length; you are given distance of the pullback in the x direction.  So you know the x and y components of the vector.)

Assuming that the pendulum is pulled back in the positive x direction, the pendulum string runs from the origin up and to the left to the point (-10 cm, 200 cm).  The direction of a vector running along the string, measured counterclockwise from the positive x axis, is therefore

• angle = arcTan(200 cm / (-10 cm) ) = 2.9 deg + 90 deg = 92.9 deg.

What is the direction of the tension force exerted on the mass?

The tension force on the mass is directed along the string, at angle of 92.9 deg as measured counterclockwise from the positive x axis.

What therefore are the horizontal and vertical components of the tension?

The tension is 5 N and its components are

• T_x = 5 N * cos(92.9 deg) = .25 N and

• T_y = 5 N * sin(92.9 deg) = 5 N, approx.

What therefore is the weight of the pendulum, and what it its mass? 

Since the pendulum is in equilibrium, its weight is the same as the vertical component of the tension, which is about 5 N.  Its mass is therefore

• m = weight / g = 5 N / (9.8 m/s^2) = .51 kg.

What is its acceleration at this instant? 

The net force on the pendulum is the x component of the tension, which is about .25 N, so its acceleration is

• a = F_net / m = .25 N / (.51 kg) = .49 m/s^2, approx.. 

Note that this is about 1/20 the acceleration of gravity, and that the 10 cm x displacement is about 1/20 of the 2 meter length.