# A Stone of Mass 0.25 Kg Tied to the End of a String is Whirled Round in a Circle of Radius 1.5 M with a Speed of 40 Rev./Min in a Horizontal Plane. What is the Tension in the String? What is the Maximum Speed with Which the Stone Can Be Whirled Around If the String Can Withstand a Maximum Tension of 200 N? - Physics

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

#### Solution 1

Mass of the stone, m = 0.25 kg

Radius of the circle, r = 1.5 m

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

Angular velocity , omega= v/r = 2pin

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

T = F_"Centripetal"

= (mv^2)/r = mromega^2 = mr(2pin)^2

= 0.25 xx 1.5 xx (2xx3.14xx2/3)^2

= 6.57 N

Maximum tension in the string, Tmax = 200 N

T_"max" = (mv_"  max"^2)/m

:.v_"max" = sqrt((T_max xx r)/m)

 = sqrt((200xx1.5)/0.25)

= sqrt(1200) = 34.64 m/s

Therefore the maximum speed of the stone is  34 .64 m/s

#### Solution 2

Her m = 0.25 kg, r = 1.5 m

n = 40 "rpm" = 40/60 rps = 2/3 "rps"

Now T = mromega^2 = mr(2pin)^2 = 4pi^2 mrn^2

T = 4xx22/7xx22x7xx 0.25xx1.5xx(2/3)^3 = 6.6 N

if T_"max" = 200 N then from

T_"max" = mv_"max"^2 => v_"max"^2 = (T_"max" xx r)/m

or v_"max"^2 = (200xx1.5)/0.25 = 1200  => v_max  = sqrt1200 = 34.6 ms^(-1)

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 5 Laws of Motion
Q 21 | Page 111