Physics with Calc 1 9.A stone at the end of a sling is whirled in a vertical cir
ID: 1434134 • Letter: P
Question
Physics with Calc 1
9.A stone at the end of a sling is whirled in a vertical circle of radius 1.40 m at a constant speed v0 = 1.40 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
toward the center of the circle
in the direction of v0
downward
upward
(d) What is the acceleration of the stone just after it is released at A?
Magnitude (absolute value)
(?) Direction
upward toward the center of the circle
or
in the direction of v0
or
downward
Explanation / Answer
a) height of stone from ground = hi + rsin30
h = 1.50 + (1.40 sin30) = 2.2 m
to reach the ground,
vertical displacement, y = - 2.2 m
vertical initial velocity = 1.40sin60 = 1.212 m/s
a = -9.8 m/s^2
y = ut + at^2 /2
-2.2 = 1.212t - 4.9t^2
4.9t^2 - 1.212t - 2.2 = 0
t = 0.805 s
R = v cos60 t = 1.4 x cos60 x 0.805 = 0.5635 m
b)
to reach the ground,
vertical displacement, y = - 2.2 m
vertical initial velocity = - 1.40sin60 = - 1.212 m/s
a = -9.8 m/s^2
y = ut + at^2 /2
-2.2 =- 1.212t - 4.9t^2
4.9t^2 + 1.212t - 2.2 = 0
t = 0.558 s
R = v cos60 t = 1.4 x cos60 x 0.558 = 0.39 m
c) at A:
accelration before release have is radial acceleration
a_c = v^2 / r = 1.4^2 / 1.40 = 1.4 m/s^2
toward the centre.
d) after release, stone is in influence of gravity.
a = 9.8 m/s^2
direction -> downward