Question

If a 0.15 kg ball on the end of a string is swung in a vertical circle of radius .6 meters and makes 2 revolutions per second, what is the tension in the string at the very top of the circle? What is the tension in the string at the bottom of the circle? (12.7 N; 15.7 N)

Answer 1

Answer:

12.7 N

15.7 N

Explanation:

mass (m) = 0.15 kg

radius (r) = 0.6 m

speed  = 2 rps = 2 x 60 = 120 rpm

acceleration due to gravity (g) = 9.8 m/s^{2}

find the tension at the top and bottom of the circle.

Tension at the top T = $\frac{mv^{2} }{r} - mg$

• where v = speed in m/s

        v =  radius x rpm x 0.10472 = 0.6 x 120 x 0.10472 = 7.54 m/s

• we can now substitute the value of v into T = $\frac{mv^{2} }{r} - mg$

         T = $\frac{0.15x7.54^{2} }{0.6} - (0.15x9.8)$ = 12.7 N      

Tension at the bottom T' = $\frac{mv^{2} }{r} + mg$

• where v = speed in m/s

        v =  radius x rpm x 0.10472 = 0.6 x 120 x 0.10472 = 7.54 m/s

• we can now substitute the value of v into T' = $\frac{mv^{2} }{r} + mg$

         T' = $\frac{0.15x7.54^{2} }{0.6} + (0.15x9.8)$ = 15.7 N