Answer:
The speed of the stone will be 8.48 m/s.
Explanation:
When the stone is moving in the horizontal circle the centripetal force will be:
F1 = (mv^2)/r .………. (i)
In the vertical circle,
F2 = mg + (mv^2)/r ..……… (ii)
As in the vertical circle tension is 0.15 % greater so,
F2 = (1+0.15)F1 ……… (iii)
Put equation (i) and (ii) in (iii) ,
mg + (mv^2)/r = 1.15 (mv^2)/r
mg = 1.15 (mv^2)/r - (mv^2)/r
mg = 0.15 (mv^2)/r
v = √(9.8×1.10)/0.15
v = 8.48 m/s
Speed is defined as the rate of change of the distance traveled.The speed of the stone is 8.48 m/sec.
Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. It's unit m/sec
It is denoted by u for the initial speed while v for the final speed. its SI unit is m/sec.
The given data in the problem is;
r is the length=1.10 m
In the vertical case, the maximum tension in the string is 15.0% larger than the tension
v is the speed of stone=?
The stone is moving in the horizontal circle the centripetal force occurs:
[tex]\rm F_c = \frac{mv^2}{r}[/tex]
When the stone moves in a vertical circle;
[tex]\rm F_v = mg + \frac{(mv^2)}{r}[/tex]
According to the problem conditions the vertical circle tension is 0.15 % greater ;
[tex]F_V = (1+0.15)F_c[/tex]
On putting all the values then;
[tex]\rm mg + \frac{(mv^2)}{r} = 1.15\frac{(mv^2)}{r}\\\\ \rm mg = 1.15 \frac{(mv^2)}{r} - \frac{(mv^2)}{r} \\\\ mg = 0.15 (mv^2)/r\\\\ \rm v = \sqrt{9.8\times \frac{1.10}{0.15} } \\\\ \rm v = 8.48 m/s[/tex]
Hence the speed of the stone is 8.48 m/sec.
To learn more about the speed refer to the link;
https://brainly.com/question/7359669
