# A 240 g block on a 60 cm long string swings in circle on a horizontal, frictionless table at 95 rpm. What is the speed of the block?What is the tension of the string?

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> Let,

m = 240 g = 0.24 kg

L = 60 cm = 0.6 m

> Express the given angular velocity in radian/second ( rad/s).

`omega = 95 (rev)/min * (2pi rad)/(1 rev) *( 1 min)/(60 s)= 19/6 pi (rad)/s`

> Determine the weight of the block.

`W = mg = (0.24 kg)( (9.81 m/s) = 2.35 N`

> To solve for the tension (T) in the spring, consider the horizontal components of the forces acting on the block.

`sumF_x = 0`

`CF - Tcos theta = 0`

where, CF is the centrifugal force

`CF = (W/g )v^2/R = m v^2/R= m (omega R)^2/R = m omega^2R`

` cos theta = R/L`

* (Note that R is the raduis of the circle formed by the rotating string.) *

So we have,

`CF - T cos theta = 0`

`m omega^2 R - T(R/L) = 0`

`m omega^2R = T(R/L)`

>Cancel the R's on both sides of the equation.

`m omega^2 = T/L`

>Substitute the values of m, `omega` and L.

`0.24 kg (19/6 pi (rad)/s )^2 = T / (0.6m)`

`0.24 kg ( 0.6m) (19/6 pi (rad)/s)^2 = T`

`1.44 pi^2 N = T`

`14.21 N = T`

Hence, Tension in the spring is 14.21 N.

To solve for the speed of the block, we need to consider the vertical components of the forces acting on the block.

`sumF_y = 0`

`T sin theta - W = 0`

`Tsin theta = W`

Substitute the values of T and W. Then solve for `theta` .

`theta = sin^-1 (W/T) = sin ^ -1 = (2.35 N )/ (14.21N) `

`theta = 9.51 deg`

Then, solve for R.

`cos theta = R/L`

`R = (cos theta ) L `

Substitute the values of L and the angle.

`R = ( cos 9.52) 0.6 =0.59 m `

``So the speed of the block is:

`v = omega r = 19/6 pi (rad)/s * .59 m = 5.9 m/s`

This is about angular motion. Since there is no friction on the table frictional force is 0 hence no speed reduction from friction.

Speed of the block (V) = r*w where r is radius of the circle in meters and w is the angular speed in radians per second.

r = 60 cm = 0.6m

w = 95rpm = 95*2*pi/60

So speed of the block (V) = 0.6*95*2*pi/60

= 5.969 m/s

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