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a 250-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.70m. the ball makes 2.0 revolutions per second. (a) what is its centripetal acceleration? (b) what force is to be exerted on the string?
given: m=250g=0.25kg; r=0.70m; =2.0rps (a) the given angular velocity is in evolution per second, we need to convert it yet to radian measures


Sagot :

Answer:

1. The centripetal acceleration

[tex] a_c \approx \boxed {110.28 \, \text{m/s}^2}[/tex]

2. The force exerted on the string

[tex]F \approx \boxed {27.57 \, \: {N}}[/tex]

Given Data:

• Mass of the ball,

[tex]m = 0.25 \, \: {kg}[/tex]

• Radius of the circle,

[tex]r = 0.70 \, \: {m}[/tex]

• Revolutions per second,

[tex]f = 2.0 \, \: {rps}[/tex]

(a) Centripetal Acceleration:

Substituting;

[tex]f = 2.0 \, \: {rps}[/tex]

[tex]\omega = 2\pi \times 2.0 = 4\pi \, \text{rad/s}[/tex]

The centripetal acceleration ( a_c ) is given by:

[tex]a_c = \omega^2 r[/tex]

Substituting:

[tex] \omega = 4\pi[/tex]

[tex]r = 0.70 \, \: {m}[/tex]

[tex]a_c = (4\pi)^2 \times 0.70[/tex]

[tex]a_c = 16\pi^2 \times 0.70[/tex]

[tex]a_c \approx 16 \times 9.87 \times 0.70[/tex]

[tex]a_c \approx 110.28 \, \: {m/s}^2 [/tex]

(b) Force on the String:

The force exerted on the string is the centripetal force, which can be calculated using:

[tex]F = m a_c[/tex]

Substituting

[tex]m = 0.25 \, \text{kg}[/tex]

[tex] a_c = 110.28 \, \: {m/s}^2 \):[/tex]

[tex]F = 0.25 \times 110.28[/tex]

[tex]F \approx \boxed{ 27.57 \, \text{N}}[/tex]