Question

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 200N?

Answer 1

Hey mate,

# Answer-
a) T = 6.57 N
b) vmax = 34.64 m/s

# Explanation-
m = 0.25 kg
r = 1.5 m
f = 40 rpm = 2/3 Hz
Tmax = 200 N

# Solution-
a) Tension in the string-
Here, centripetal force is provided by tension in the string.
Hence,
T = Fc
T = mrω^2
T = 0.25×1.5×(2×3.14×2/3)^2
T = 6.57 N

Tension in the string is 6.57 N.

b) Maximum speed-
At Tmax, velocity is Vmax,
Tmax = m(vmax)^2/r
200 = 0.25(vmax)^2/1.5
vmax^2 = 200×1.5/0.25
vmax^2 = 1200
vmax = 34.64 m/s

Maximum speed of stone is 34.64 m/s.

Hope that was useful...

Answer 2

$\huge\fcolorbox{black}{teal}{Solution:-}$

✏️Answer:

★ Tension = 6.58 N

★ Maximum speed = 34.64 m/s

✏️Explanation:

★ Given:

•Mass of the stone = 0.25 kg

•Radius of the circular path = 1.5 m

• Frequency = 40 rev/min

•Maximum tension = 200 N

★ To find:

•Tension in the string [Maximum speed with which the stone can be whirled around if the maximum tension it can withstand is 200 N]

★ Rationale:

• The tension produced in the string is given by the centripetal force which is given by,

T = m v²/r

T = m r ω² (∵ ω = v/r)

where m is the mass of stone,

r is the radius

ω is the angular velocity

First finding the angular velocity of the stone ω

ω = 2 π f

where f is the frequency,

Substitute the data,

ω = 2 × π × 40 = 80 π rad/min = 4.189 rad/s

Hence,

T = 0.25 × 1.5 × 4.189²

T = 6.58 N

Hence the tension produced in the string is 6.58 N

We know that maximum tension is given by,

Tmax = mv²max /r vmax

= √Tmax x r/m

= √200 x 1.5/0.25

= √1200

= 34.64 m/s

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

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I hope this helps! :)