Question

A certain string can withstand a maximum tension of 40 N without breaking. A child ties a 0.37kg stone to one end and, holding the other end, whirls the stone in a vertical circular radius 0.91 m, slowly increasing the speed until the stone breaks. (a) Where is the stone on its path when the stone breaks? (b) What is the speed of the stone as the string breaks?

Answer 1

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= $\sqrt{r(\frac{T}{m}-g) }$

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 2

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