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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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