Question

a 1.13 kg ball is swung vertically from a 0.50 m cord in uniform circular motion at a speed of 2.4 m/s. what is the tension in the cord at the bottom of the balls motion

Answer 1

24 N

Step-by-step explanation:

Draw a free body diagram of the ball. There are two forces:

Weight force mg pulling down.

Tension force T pulling up.

Sum the forces towards the center of the circle:

∑F = ma

T − mg = m v² / r

T = m (g + v² / r)

Given m = 1.13 kg, g = 9.8 m/s², v = 2.4 m/s, and r = 0.50 m:

T = (1.13 kg) (9.8 m/s² + (2.4 m/s)² / 0.50 m)

T = 24 N

Answer 2

The tension in the cord is 24.1 Newton

Step-by-step explanation:

Step 1: Data given

mass of the ball = 1.13 kg

Cord = 0.50m

velocity = 2.4 m/s

Step 2: Calculate the tension

Fnet = F(c)

T - F(g) = F(c)

⇒ with F(g) = force of gravity = m*g

⇒ with F(C) = centripetal force = (mv²)/r

T = (mv²)/r + mg

⇒ with m = mass = 1.133 kg

⇒ with v = velocity = 2.4 m/s

⇒ with r = radius = 0.50 m

T = (1.13 * 2.4²) / 0.50 + 1.13 * 9.81

T = 24.1 Newton

The tension in the cord is 24.1 Newton