Answer:
The tension in the rod as the ball moves through the bottom circle is 9.8 N
Explanation:
When the ball is released from rest, the centripetal force equals the weight of the ball. So mv²/r = mg where m = mass of ball = 0.5 kg, v = speed of ball, r = radius of vertical circle = length of rod = 0.5 m and g = acceleration due to gravity = 9.8 m/s²
v = √gr = √9.8 m/s² × 0.5 m = √4.9 = 2.21 m/s
Now at the bottom of the circle T - mg = mv²/r where T = tension in the rod
T = m(g + v²/r)
= m(g + (√gr)²/r)
= m(g+ gr/r)
= m(g + g)
= 2mg
= 2 × 0.5 kg × 9.8 m/s²
= 9.8 N
So, the tension in the rod as the ball moves through the bottom circle is 9.8 N
Answer:
18.8m
Explanation:
This implies calculating the relative speed first then find the distance d
So
relative speed = 2.75- 1.25 = 1.5 m/s
So assume the distance between them to be say d
so, d/1.5= 12.5
d = 18.8m
B.
increases as the tension of the string increases
Answer:
10seconds
Explanation:
use the formula a=v final_v inital/time
Answer:
Explanation:
The shear stress due to torque can be calculed by using the following model:
The maximum torque on the section is:
The Torsion Constant for the circular tube is:
Now, the require output is computed: