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This diagram shows a rubber bung, of mass 200 g, on the end of a length of string being swung in a horizontal circle of radius 40 cm. The string makes an angle of 56° with the vertical.
Calculate:
i the tension in the string
ii the angular speed of the bung
iii the time it takes to make one complete revolution​

Respuesta :

I) Tension in the spring = 2N

ii) Angular speed of the bung=5 rads-¹

iii) The time(T) it takes to make one complete revolution=1.256 seconds

Circular motion

Formula used to calculate the time(T) it takes to make one complete revolution = 2π √(l/g)

l = radius= 40cm = 0.4m

g = acceleration due to gravity = 10m/s-²

π =3.14( constant)

T = 2× 3.14 × √0.4/10

T = 6.28 ×√0.04

T = 6.28 × 0.2

T= 1.256 seconds

Formula to calculate angular speed(w)= 2π/T

where T= 1.256 secs

Therefore w = 2×3.14/1.256

w = 6.28/ 1.256

w = 5 rads-¹

Formula to calculate Tension in the spring(F)

= mrw²

where m= 200 g = 0.2kg

r = radius= 40cm = 0.4m

w = angular speed = 5 rads-¹

Therefore F = 0.2 × 0.4 × 5²

= 0.08 ×25

= 2N

Learn more about angular speed here:

https://brainly.com/question/6860269

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