A stone of mass 0.1 kg tied to one end of a string 1.0 m long is revolved in a horizontal circle at the rate of 10 / pie revolutions per sec . Calculate the tension of the string

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Solution

$D e a r S t u d e n t, V = r ω V = 2 π n \times r = 1 \times 2 \times π \times \frac{10}{π} v = 20 m / s s o a = r ω^{2} a = 1 \times {(2 π n)}^{2} a = 400 m / s^{2} T e n s i o n T = m r ω^{2} = 400 N R e g a r d s$

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A stone of mass 0.1 kg tied to one end of a string 1.0 m long is revolved in a horizontal circle at the rate of 10 / pie revolutions per sec . Calculate the tension of the string

Dear Student , V=rωV=2πn×r=1×2×π×10πv=20m/sso a=rω2a=1×2πn2a=400m/s2Tension T=mrω2=400NRegards

$D e a r S t u d e n t, V = r ω V = 2 π n \times r = 1 \times 2 \times π \times \frac{10}{π} v = 20 m / s s o a = r ω^{2} a = 1 \times {(2 π n)}^{2} a = 400 m / s^{2} T e n s i o n T = m r ω^{2} = 400 N R e g a r d s$