Question

There is a small hole in a table A string of length 1 m passes through it . Two bodies of masses 70 g and 100 g are attached at the ends . The 100 g mass hangs freely at a depth in a circle with frequency equals to

Answer 1

Given info : There is a small hole in a table. a string of length 1 m passes through it. Two bodies of masses 70g and 100g are attached at the ends.

To find : The 100g hangs freely at a depth of 60cm from the table. If this mass is to be in equilibrium, the other mass should rotate in a circle with a frequency equal to......

Solution : tension in the string attached 100g mass, T = mg = 100 × 10¯³ kg × 10 m/s² = 1 N

At equilibrium,

Tension in the string = centripetal force acting on mass of 70g

⇒T = mω²r

Here, r = 100 - 60 = 40 cm = 0.4 m

m = 70 × 10¯³ kg

⇒1 N = (70 × 10¯³ × ω² × 0.4 m)

⇒ω² = 1000/28

⇒ω = 250/7

⇒2πf = √(250/7) = 5.98

⇒f = 5.98/2π Hz

Therefore the frequency equals to 5.98/2π Hz

Answer 2

Answer:

root 140/ 4π is answer dear