# What is the tension in units of N of the string in the scenario below if the acceleration of gravity is 9.8 m/s^2?An air puck of mass o.257 is tied to a string and allowed to revolve in a circle...

What is the tension in units of N of the string in the scenario below if the acceleration of gravity is 9.8 m/s^2?

An air puck of mass o.257 is tied to a string and allowed to revolve in a circle of radius 0.78 m on a horizontal, frictionless table.

The other end of the string passes through a hole in the center of the table and a mass of 1.51 kg is tied to it. The suspended mass remains in equilibrium while the puck revolves.

*print*Print*list*Cite

### 2 Answers

The centripetal force of a rotating body is given by mv^2/r. Where m is the mass of the body, r is the radius of the circle it is rotating in and v is the velocity it is moving at.

Here we have the mass = 0.257 kg (I assume it is kg), the radius 0.78 m, but the velocity is not specified. Let's assume it is V.

Now the centripetal force is due to the mass tied at the end of the string and is equal to weight of the body or 1.51*9.8 N.

So we have 0.257*V^2/0.78 = 1.51*9.8

=> V^2 = 1.51*9.8*7.8/0.257

=> V^2 = 449.1

=> V = 21.19 m/s

So the puck rotates at 21.19 m/s

**The force due to weight of the body is the tension in the string, so the tension is equal to 9.8*1.51 = 14.798 N**

When the air puck moves around in a circle it tries to move away, where as the weight force at the centre pulls it in a reverse direction. Thus system is in equlibrium .

TheforeĀ Tension force , T = massĀ tied to the string* acceleration due to gravity.

T = 1.51kg*9.8

T = 1.14.798 newton

Therefore the tension in the string is 1.14798 N.