Question

A stone weighing 0.4 kg is whirled around in a horizontal
circle at the end of a string 0.5 m long with a speed of 180
rpm. Calculate the tension in the string.

pls guys it's bit urgent so pls answer fast properly on a notebook...

Answer 1

Answer:

tension =0.4/0.5*180

=0.4*3.6

=4*36

=1.44

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Answer 2

Answer:

The tension in the string is 70.98 N

Explanation:

In the question mass of stone is given which is 0.4 kg. Radius is 0.5 m and speed is 180 rpm.

Tension in the string will be equal to its centripetal force.

First covert 180 rpm into 180 rps

Since 1 rpm is equal to 2 π / 60  

Therefore, 180 rpm = 180 x 2 x 3.14/60 = 18.84 r/sec

Tension = centripetal force = $\bold{\mathrm{m} \cdot \mathrm{r} \cdot \omega^{2}}$

= 0.4 x 0.5 x (18.84)^{2}

= 0.4 x 0.5 x 354.94

Tension = 70.98 N