A stone at the end of a sling is whirled in a vertical circle of radius 1.40 m a

Question

A stone at the end of a sling is whirled in a vertical circle of radius 1.40 m at a constant speed v0 = 2.50 m/s as in Figure P4.57. The center of the string is 1.50 m above the ground.

(a) What is the range of the stone if it is released when the sling is inclined at 30.0° with the horizontal at A?

(b) What is the range of the stone if it is released when the sling is inclined at 30.0° with the horizontal at B?

(c) What is the acceleration of the stone just before it is released at A?
Magnitude (absolute value)
Direction

in the direction of v0toward the center of the circle    upward downward

(d) What is the acceleration of the stone just after it is released at A?
Magnitude (absolute value)
Direction

in the direction of v0toward the center of the circle    upward downward

Explanation / Answer

Newton's II law => F = ma = F(resultant)

this F(resultant) provides the centripetal force necessary to keep the object in a circular path = mv²/r

constant speed = 2.50 m/s

c) just before release equation is (along the circular path radii)

T'' + mgsin = mv²/r

[please do note that since this is uniform vertical circular motion the tension and therefore the string cannot always be along the radius of the circular path since that would leave the tangential component of weight = mg unbalanced. T'' is the component of T along the radii and the other tangential component will balance mgcos to ensure there is no linear acc.]

since it is at a constant speed there is no linear acceleration (tangential)

therefore only acc. is radial acceleration = a = v²/r = 2.5² / 1.4 = 4.46 m/s²

d) after release the object is a projectile and only acceleration is g = 9.81 m/s² vertically downwards


a) total vertical distance it descends = 1.5 + 1.4*sin30° = 2.2 m

u = 2.50*sin60° upwards = -2.165 m/s

s = ut + ½gt² => 2.2 = -2.165t + 4.905t² =>

4.905t² - 2.165t - 2.2 = 0 => t = 0.925 s

range = 2.50cos60°t = 1.25x0.925 ~= 1.156 m


b) u = 2.50*cos30° downwards = 2.165 m/s

s = ut + ½gt² => 2.2 = 2.165t + 4.905t² =>

4.905t² + 2.165t - 2.2 = 0 => t = 0.48 s

range = 2.50*sin30°t = 1.25x0.48 ~= 0.6 m

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