Q4. An Olympic diver weighs 70 kg. Prior to competition, he lays on a balance bo

Question

Q4. An Olympic diver weighs 70 kg. Prior to competition, he lays on a balance board so his COM position can be computed. He is 181 cm tall, and a scale at his head (W2) reads 64 lbs. The length of the board is 8 ft (see diagram).Please show your work!!!

a) What is the height of his COM relative to his feet (in cm)? (2 pts)

b) What is his potential energy when he is standing on the pool deck? (2 pts)

c) He climbs to the top of a 10 m platform. What is his potential energy as he stands on the edge of the platform? (2 pts)

d) As he performs his dive, he jumps upward and displaces his COM 10 more cm upwards. When his CM is at a height of 5 m from the surface of the water, how much kinetic energy does he possess? (2 pts)

e) Just before he enters the water, is COM is 1.2 m above the surface. How fast is his COM travelling? (2 pts)

An Olympic diver weighs 70 kg. Prior to competition, he lays on a balance board so his COM position can be computed. He is 181 cm tall, and a scale at his head (W2) reads 64 lbs. The length of the board is 8 ft (see diagram).Please show your work!!! a) What is the height of his COM relative to his feet (in cm)? What is his potential energy when he is standing on the pool deck? He climbs to the top of a 10 m platform. What is his potential energy as he stands on the edge of the platform? As he performs his dive, he jumps upward and displaces his COM 10 more cm upwards. When his CM is at a height of 5 m from the surface of the water, how much kinetic energy does he possess? Just before he enters the water, is COM is 1.2 m above the surface. How fast is his COM travelling?

Explanation / Answer

Weight at head = 64 lbs = 29.03 Kg

W1 + W2 = 70

W1 = 70-29.03 = 40.97 Kg

Apply torque balance

Let distance of her feet from C.O.M. is x

W1*g*x = W2*g*(L-x) where L = 8 ft = 2.438 m

40.97*g*x = 29.03*(2.438-x)*g

x = 1.011 m = 101.1 cm

(a)

Hieght of C.O.M. relative to his feet = 101 cm

(b)

Potential energy = m*g*h = 70*9.8*1.011 = 693.55 J

Where h is hieght of C.O.M. from earth.

(c)

Potential energy depends on the hieght of C.O.M. from earth.At surface of earth it is zero.

So after getting on a board 10m hiegher we get h = 10+1.011 = 11.011 m

So potential energy then is m*g*h = 70*9.8*11.011 = 7553.55 J

(d)

When he raises 10cm = 0.1 m more the hieght of C.O.M. becomes = 11.011+0.1 = 11.111m

So at maximum hieght potential energy P.E.1 = m*g*h = 70*9.8*11.111 = 7622.15 J

At maximum hieght velocity is zero .So kinetic energy is zero.

Apply energy conservation

P.E1 + K.E.1 at top most point = P.E.2 + K.E.2 When he is at 5 m.........(Equation 1)

P.E.2 at 5 m = m*g*5 = 3430 J

Put values in equation 1

7622.15 + 0 = 3430 + K.E.2

Kinetic energy = 4192.15 J

(e)

Potential energy when he is 1.2 m above water =P.E.3 = m*g*h = 70*9.8*1.2 = 823.2 J

Apply ENERGY CONSERVATION

P.E.1 + K.E.1 = P.E.3 + K.E.3

7622.15 + 0 = 823.2 + K.E.3

K.E.3 = 6798.95 = 6799 J

1/2*m*v^2 = K.E.3 = 6799

v = (2*6799/70)^1/2 = 13.94 m/s

Related Questions

Navigate

• Browse (All)
• Browse Q
• Subjects

---

---

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