A string of cross-sectional area 4 mm² and length 0.5 m is connected to a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5 m. The body acquires a speed of 5 m s⁻¹ at the bottom of the circular path. The strain produced in the string when the body is at the bott…

Question

A string of cross-sectional area 4 mm² and length 0.5 m is connected to a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5 m. The body acquires a speed of 5 m s⁻¹ at the bottom of the circular path. The strain produced in the string when the body is at the bottom of the circle is ×10⁻⁵. (Use Young's modulus =10¹¹ N m⁻² and g=10 m s⁻²)

Accepted Answer

30. Given: A=4 mm²=4×10⁻⁶ m², m=2 kg, r=0.5 m, v=5 m s⁻¹, Y=10¹¹ N m⁻². At the lowest point the tension pulls up while the weight pulls down, and their difference supplies the centripetal force: T-mg=(mv²)/r. T=mg+(mv²)/r=2×10+(2×25)/(0.5)=20+100=120 N. Strain =T/(AY)=(120)/(4×10⁻⁶×10¹¹)=(120)/(4×10⁵)=3×10⁻⁴. Written as a multiple of 10⁻⁵ this is 30×10⁻⁵, so the answer is 30.

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