4. (30 pts.) A child of mass 30 kg sits on top of a uniform toy tower of mass 70

Question

4. (30 pts.) A child of mass 30 kg sits on top of a uniform toy tower of mass 70 kg and length 3.0 m. The tower gets tilted due to the child's smart move. You heroically reach the tower to save the child, and you manage to stop the tower when it makes an angle of 30 degrees from vertical. (a) If you apply a force perpendicular to the tower at 1.6 m from the bottom of the tower, how much force do you need to apply in order to keep the tower in equilibrium? 13 61 = ML2/12), what is its monnent of inertia about its (b) If we assume that the tower is a thin rod ( edge (I of just the tower)? (c) If we assume that the child is a point mass, what is its moment of inertia about the lower end of the tower (I of just the child)? (b) If you suddenly stop supporting the tower at the configuration shown in the figure, what would the speed (linear speed v, not angular speed w) of the child be when the child hits the ground? Assume that the tower and the child hit the ground at the same time

Explanation / Answer

A) Balancing torque about bottom,

F*x = mgL sin 30 degree + MgL/2 sin 30 degree

F*1.6 = 30*9.8*0.5*3 + 70*9.8*1.5*0.5 = 955.5

F = 955.5/1.6 = 597.2 N answer

B) I = 1/12 ML^2 + M(L/2)^2 = 1/3 ML^2 = 1/3*70*3^2

= 210 kgm^2

C) I = mL^2 = 30*3^2 = 270 kgm^2

D) 0.5 i w^2 = decrease in potential energy = mgL + MgL/2

0.5*(210+270)w^2 = 30*9.8*3+70*9.8*1.5 =1911 J

w = sqrt (2*1911/480)

= 2.82 rad/s

v = wL = 2.82*3

= 8.46 m/s answer

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.