Question

A stone of mass 100 gm is whirled in a
horizontal circle at the end of a string 50
cm long. The string can withstand a
maximum tension of 288 dynes. At what
maximum speed can the stone be
whirled without stopping the string?​

Answer 1

Explanation:

Mass of the stone, m=0.25 kg

Radius of the circle, r=1.5 m

Number of revolution per second, n=40/60=2/3rps

Angular velocity, ω=2πn

The centripetal force for the stone is provided by the tension T, in the string, i.e.,

T=mω

2

r

=0.25×1.5×(2×3.14×(2/3))

2

=6.57 N

Maximum tension in the string, T

max

=200 N

T

m

ax=

r

mv

max

2

v

max

=(

m

T

max

r

)

1/2

=(200× 1.5/0.25)

1/2

=(1200)

1/2

= 34.64 m/s

Therefore, the maximum speed of the stone is 34.64 m/s.

Answer 2

The maximum speed from that the stone can be whirled without stopping the string is 12m/sec.

Given:-

Mass of stone = 100gm

Length of the string = 50cm

Maximum tension of string = 288dynes

To Find:-

The maximum speed from that the stone can be

whirled without stopping the string.

Solution:-

We can easily find out the maximum speed from that the stone can be whirled without stopping the string by using these simple steps.

As

Mass of stone (m) = 100gm

Length of the string (r) = 50cm

Maximum tension of string (t) = 288dynes

Maximum speed (v) =?

Here since all the units are in CGS system, so there is no need to change any of the following dimensions.

According to the formula,

$t = \frac{m {v}^{2} }{r}$

${v}^{2} = \frac{t \times r}{m}$

on putting the values,

${v}^{2} = \frac{288 \times 50}{100}$

${v}^{2} = \frac{288}{2}$

${v}^{2} = 144$

$v = \sqrt{144}$

$v = 12$

Hence, The maximum speed from that the stone can be whirled without stopping the string is 12m/sec.

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