Answer :
Answer:
a. 11.96 m/s b. 43,014 N
Explanation:
a. At the top of the circle, the centrifugal force must equal the normal force on the vertical = weight of motorcycle + weight of physics major.
(m₁ + m₂)v²/r = (m₁ + m₂)g
Its speed is thus v = √gr r = radius of vertical circle = 14.6 m
v = √gr = √(9.8 × 14.6) = √143.08 = 11.96 m/s
b. The normal force at the bottom equal the centripetal force + the weight of the motorcycle + weight of physics major
N = (m₁ + m₂)v²/r + (m₁ + m₂)g = (m₁ + m₂)gv²/r
m₁ = mass of physics major = 72.0 kg
m₂ = mass of motorcycle = 40.0 kg
v = 2 × 11.96 m/s = 23.92 m/s (since it is twice that at the top of the circle)
r = radius of circle = 14.6 m
N = (m₁ + m₂)gv²/r = (72.0 + 40.0) × 9.8 × 23.92²/14.6 = 112 × 9.8 × 23.92²/14.6 = 43,014.37 N = 43.014 kN
Answer:
A) 11.968 m/s
B) 2668.4N
Explanation:
Mass of motorcycle (Mm) = 40kg
Mass of Physics major (Mp) = 72kg
Radius (r) = 14.6m
A) Using Newton's second law of motion, the net force at the top of circle is given by;
ΣF = - Ma(c) = -Mv²/r
And so; Resultant(R) = Mg - Ma(c)
Now, R= 0 since it loses contact.
Thus,
Mg - Ma(c) = 0
From above, -Ma(c) = -mv²/r
So we'll get; Mg - M(vmin)²/r = 0
Thus,Mg = M(vmin)²/r
So, g = (vmin)²/r
(vmin)² = rg
Vmin = √rg
Vmin = √(14.6 x 9.81)
Vmin = √143.226 = 11.968 m/s
B) From the earlier equation, the net force acting on the motorcycle is;
R(m) - M(m)g - M(p)g = M(m)a(c) = (M(m)v²)/r
Thus; let's use;
R(m) - M(m)g - M(p)g = (M(m)v²)/r
So;
R(m) = M(m)g + M(p)g + (M(m)v²)/r
R(m) = g[M(m) + M(p)] + (M(m)v²)/r
From the question, v is now double.
So, v = 2 x 11.968 = 23.936
R(m) = 9.81[40 + 72] + [(40 x 23
936²)/14.6]
R(m) = 1098.72 + 1569.68 = 2668.4N