Question

Please answer #3 and show your work. I will rate your answer.

h в о h, 3. A sphere is rolling without slipping at a (center-of-mass) speed of vA = 5 m/s along a horizontal surface. It encounters a ramp of height h1 = 0.5 m that redirects its motion vertically. The sphere never slips while in contact with the surface. (a) What is the sphere's speed as it leaves the surface at point B? (b) What maximum height (h2 in the figure) do es it reach? DO Co 4. A rod of length 2R is dropped on a freely rotating cylindrical turntable of radius R so that it lies as a diameter on the turntable. If both turntable and rod have the same mass m, and the turntable is initially rotating at angular speed wo, what is the new angular speed once the two bodies rotate in unison?

Answer 1

3)

a)

v_A = initial speed of the sphere at A = 5 m/s

w_A = angular speed of sphere at A

r = radius of the sphere

m = mass of the sphere

moment of inertia of the sphere is given as

I = (0.4) m r²

v_B = final speed of the sphere at B = ?

w_B = angular speed of sphere at B

h₁ = height gained from A to B

using conservation of energy between A and B

Total energy at A = Total energy at B

(0.5) m v_A² + (0.5) I w_A² = (0.5) m v_B² + (0.5) I w_B² + m g h₁

(0.5) m v_A² + (0.5) (0.4) (mr²) (v_A/r)² = (0.5) m v_B² + (0.5) (0.4) (mr²) (v_B/r)² + m g h₁

(0.5) m v_A² + (0.2) m v_A² = (0.5) m v_B² + (0.2) m v_B²+ m g h₁

(0.7) v_A² = (0.7) v_B² + g h₁

(0.7) (5)² = (0.7) v_B² + (9.8) (0.5)

v_B = 4.24 m/s

b)

w_C= angular speed at C = w_B = angular speed at B

Using conservation of energy from point B to point C

Kinetic energy at B = rotational kinetic energy at C + Potential energy at C

(0.7) m v_B² = (0.5) I w_C² + m g h₂

(0.7) m v_B² = (0.5) (0.4) (mr²) (v_B/r)² + m g h₂

(0.7) m v_B² = (0.2) m v_B² + m g h₂

(0.5) v_B² = g h₂

(0.5) (4.24)² = 9.8 h₂

h₂ = 0.92 m