Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A stone at the end of a sling is whirled in a vertical circle of radius 1.10 m a

ID: 1429618 • Letter: A

Question

A stone at the end of a sling is whirled in a vertical circle of radius 1.10 m at a constant speed v0 = 1.30 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 v0, downward, upward, or toward the center of the circle?
(d) What is the acceleration of the stone just after it is released at A?
Magnitude (absolute value)?

Direction... is it in the direction of V0, Downward, upward, or toward the center of the circle ?

Figure P4.35 (a = 16.0 m/s2) represents the total acceleration of a particle moving clockwise in a circle of radius r = 3.00 m at a certain instant of time.


Figure P4.35

A1 S0.0° 30.0°

Explanation / Answer

vy = sin 60°*1.30 --> vy = 1.125 m/s           

vx = cos 60°*1.30 --> vx = 0.650 m/s

(a)  ay = Dvy / Dt   --> -9.8 = (0-1.125) / ttop --> ttop     = 0.115 sec

Initial vertical distance is dy-init = 1.5+sin30°*1.10 --> dy-init = 2.05 m

dtop = ½ a ttop2 + vx ttop + dy-init   --> dtop = -0.065 + 0.075 + 2.05 --> dtop = 2.06 m

dbottom = dtop = ½ a tbottom2 + vy-ini tbottom + dy-ini --> 2.06 = ½ 9.8 tbottom2 + 0 + 0 --> tbottom = 0.648 sec

ttotal = ttop + tbottom è ttotal = 0.115 + 0.648 --> ttotal = 0.763 sec

vx = x / ttotal --> 0.650 m/s = x / 0.763 sec --> x = 0.50 meters

(b)      dabove ground = dy-init = vo in y compt + ½ at2 -->           2.05 =1.3t + 4.9t2   

4.9t2 + 1.3t – 2.05 = 0 --> t = 0.53 seconds

vx = x /t    --> 0.650 = x / 0.53    --> x = 0.34 meters

(c) a = v2 / r    --> a = (1.3)2 / 1.10    --> a = 1.54 m/s2

(d) Immediately after it’s release at any point, the only force acting on it (neglecting air friction) is gravity which has a rate of acceleration of 9.8 m/s2 pointed down

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote