Question

A 10kg ball attached at the end of a rigid massless rod of length 1m
rotates at constant speed in a horizontal circle of radius 0.5m and period
of 1.58s as shown in the figure. The force exerted by the rod on the ball
is (g =10m/s)
-->10 kg

Answer 1

Step-by-step explanation:

Given A 10 kg ball attached at the end of a rigid mass less rod of length 1 m rotates at constant speed in a horizontal circle of radius 0.5 m and period  of 1.58 s as shown in the figure. The force exerted by the rod on the ball  is

• Now there is a ball and it is moving in the horizontal direction. Now we need to find the force exerted by the rod on the ball.  
• Now there is a downward force on the ball and the ball has a tendency to fall. But there is a support from the rod.
• Mass = 10 kg
• So we have F1 = mg
•                          = 10 x 10
•                          = 100 N
• Velocity = circumference / time period
•                  = 2π r / T
•                  = 2 x π x 0.5 / 1.58
•                  = 1.98
•                  = 2 m/s
• So F2 = mv^2 / r
•          = 10 x 2 x 2 / 0.5
•         = 80 N
• Now net force will be
• Force = √F1^2 + F2^2
•     = √(100)^2 + (80)^2
•      = √16400
•      = 128 N

Reference link will be

https://brainly.in/question/16701237