1.- A ball on the end of a rope is spun in a horizontal circle of 0.400 m radius

Question

1.-

A ball on the end of a rope is spun in a horizontal circle of 0.400 m radius. The plane of the circle is at 1.80 m above ground. In a moment the string breaks so the ball hits him ground at a horizontal distance of 2.50 m from the point where the rope broke. Just before the rope breaks, calculate a) the centripetal acceleration of the stone, b) the angular velocity of the stone, c) the rotation frequency and d) the period of rotation.

Replies:    a) = 42.6 m/s2 accentuated

b) = 10.3 rad / s,

c) f = 1.64 Hz, d) 0609 s.

I did, but my results are wrong, someone can help me with the solution.

2.- In a moment, a particle moving in the opposite direction to clockwise in a circle of radius r = 2.0 m has a speed v = 8.0 m / s, and acceleration at an angle  total = 60 ° with respect to the radial direction, with tangential component in the opposite direction to the direction of the velocity. a) Calculate the total particle a cceleration at that instant. b) Assuming that the angular deceleration is constant, how long to stop completely? 

Replies: a) atot = 64 m/s2, b) t = 0.144 s.

Explanation / Answer

1. a) first calculate the velocity of the stone when the rope broke

h = 1/2gt2     ===> t = sqrt(2h/g)

s = vt

where h = 1.80 m and s = 2.50 m

so v = s/t = s*sqrt(g/2h)

centripetal acceleration = v2/R = s2(g/2h)(1/R) = s2g/(2Rh)

= 2.502*9.8/(2*0.40*1.80) = 42.53 m/s2

b) = v/R = (s/R)sqrt(g/2h) = (2.50/0.40)sqrt(9.8/2*1.80) = 10.3 rad/s

c) f = /(2) = 1.64 Hz

d) T = 1/f = 0.609 s

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