Answer:
a) at the bottom
b)9.45 m/s
Explanation:
Maximum Tension 'T'= 40 N
Mass 'm' =0.37kg
vertical circular radius 'r'= 0.91m
a) at the bottom
Because the tension would be greatest at the bottom and the string will break.
Equation: T= mg + m (v²/r)---->eq(1)
b) In order to find the speed of the stone as the string breaks, we can arrange eq(1) for the velocity.
So,
v= [tex]\sqrt{r(\frac{T}{m}-g) }[/tex]
where,
v= velocity
g=gravity due to acceleration 9.8m/s2
T=tension
m=mass and r=radius
v= √0.91 x (40/0.37 - 9.8)
v= 9.45 m/s
Therefore, the speed of the stone as the string breaks is 9.45 m/s
Answer:
A) Stone will be at the bottom
B) v = 9.46 m/s
Explanation:
We are given;
Circular Radius; R = 0.91m
Tension; T = 40N
Mass of stone; m = 0.37 kg
A) In centripetal acceleration, at the top of a swing, T=mv²/r - mg.
While at the bottom of the swing, T=mv²/r + mg
Thus, from those 2 formulas it's clear that tension will have a higher value at the bottom. Thus, we can deduce that the string will definitely break at the bottom because that is the point where the tension would be the greatest.
B) the string breaks at the bottom, thus,
T=mv²/r + mg
Let's make v the subject where v is speed of stone.
Thus, v = √[(T - mg)(r/m)]
Plugging in the relevant values to get;
v = √[(40 - (0.37 x 9.8))(0.91/0.37)]
v = √89.4604
v = 9.46 m/s
