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A 0.16kg stone attached to a string of length I = 0.22m, is whirled in a horizontal circle with a constant velocity of 4.0m/s. a) Calculate the radial i.e the centripetal acceleration of the stone. b) The tension in the string while the stone is rotating. c) The horizontal force on the stone.

Respuesta :

Answer:

PART A)

[tex]a_c = 72.7 m/s^2[/tex]

PART B)

T = 11.6 N

PART C)

[tex]F_{horizontal} = 11.6 N[/tex]

Explanation:

PART A)

Centripetal acceleration is given by

[tex]a_c = \frac{v^2}{R}[/tex]

now we have

[tex]a_c = \frac{4^2}{0.22}[/tex]

[tex]a_c = 72.7 m/s^2[/tex]

PART B)

Here Tension force of string is providing centripetal force

so we can say

[tex]T = ma_c[/tex]

[tex]T = 0.16(72.7) = 11.64 N[/tex]

PART C)

Force on the stone in horizontal direction is only tension force

so here we have

[tex]F_{horizontal} = T = 11.64 N[/tex]

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