Potential Energy Additional Problems and Solutions 7

 Problem#1

A ball is tied to one end of a string. The other end of the string is held fixed. The ball is set moving around a vertical circle without friction, and with speed at the top of the circle, as in Figure 1. At what angle θ should the string be cut so that the ball will then travel through the center of the circle?

Fig.1

Answer:
Call φ = 180⁰ – θ the angle between the upward vertical and the radius to the release point. Call v_r the speed here. By conservation of energy

∆E_mech = ∆K + ∆U

0 = K_f – K_i + U_f – U_i

0 = ½ mv_r² – ½ mv_i² + mgR cosφ – mgR

v_r² + 2gR cosφ = 3gR

v_r² = 3gR – 2gR cosφ

The components of velocity at release are v_x = v_r cosφ and v_y = v_r sinφ so for the projectile motion we have

x = v_xt

R sinφ = v_rcosφt

t = Rsinφ/(v_rcosφ)           (*)

y = v_yt – ½ gt²

–Rcosφ = v_rsinφt – ½ gt² (**)

By substitution

–Rcosφ = v_rsinφ[Rsinφ/(v_rcosφ)] – ½ g[Rsinφ/(v_rcosφ)]²

–Rcosφ = Rsin²φ/(cosφ) – ½ gR²sin²φ/(v_r² cos²φ)

 –Rv_r²cos³φ = Rv_r²sin²φcosφ – ½ gR²sin²φ

½ gRsin²φ = v_r²[sin²φ cosφ + cos³φ]

gRsin²φ = 2v_r²cosφ; because [sin²φ + cos²φ = 1]

gRsin²φ = 2(3gR – 2gR cosφ)cosφ

sin²φ = 6cosφ – 4cos²φ = 1 – cos²φ

3cos²φ – 6cosφ + 1 = 0

cos φ = {6 ± [6² – 4(3)(1)]^1/2}/6

Only the – sign gives a value for cosφ that is less than one:

cos φ = 0.1835

φ = 79.4⁰

so θ = 180⁰ – 79.4⁰ = 100.6⁰ 

Problem#2
A ball whirls around in a vertical circle at the end of a string. If the total energy of the ball–Earth system remains constant, show that the tension in the string at the bottom is greater than the tension at the top by six times the weight of the ball.

Answer:

Applying Newton’s second law at the bottom (A) and top (B) of the circle gives 

∑F_A = mv_A²/R

T_A – mg = mv_A²/R (*)

and
∑F_B = mv_B²/R

T_B + mg = mv_B²/R (**)

Adding (*) and (**) these gives

T_A = T_B + 2mg + m(v_B² – v_A²)/R

Also, energy must be conserved and

0 = ∆K + ∆U

0 = ½ mv_B² – ½ mv_A² + 0 – 2mgR)

m(v_B² – v_A²)/R = 4mg

Substituting into the above equation gives

T_A = T_B + 6mg 

Problem#3
A pendulum, comprising a string of length L and a small sphere, swings in the vertical plane. The string hits a peg located a distance d below the point of suspension (Fig. 3). (a) Show that if the sphere is released from a height below that of the peg, it will return to this height after striking the peg. (b) Show that if the pendulum is released from the horizontal position (θ = 90°) and is to swing in a complete circle centered on the peg, then the minimum value of d must be 3L/5.

Fig.3

Answer:
(a) Energy is conserved in the swing of the pendulum, and the stationary peg does no work. So the ball’s speed does not change when the string hits or leaves the peg, and the ball swings equally high on both sides.

(b) Relative to the point of suspension,

U_i = 0 and U_f = –mg[d – (L – d)] = –mg(2d – L)

From this we find that

0 = ∆K + ∆U

0 = ½mv² – mg(2d – L)

v² = 2g(2d – L)

Also for centripetal motion,

mv²/R = mg

2g(2d – L)/(L – d) = g

4d – 2L = L – d

5d = 3L

d = 0.6L   

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