please help me with these old homework questions I got wrong and explain clearly the answer and how to get to it

Question

A wind turbine is spinning at an angular speed of 6 rad/s. Each blade of the turbine is 2m long.

What is the tangential speed of a point at the tip of a blade?

2m/s

3m/s

6m/s

12m/s

I chose 6m/s but got it wrong

A Compact disk spinning at 30 rev./min, when the disk player is turned off. It takes 2 minutes to the disk to come to a complete stop. Find the angular displacement of the disk.

30 rad

30 rev

188 rev

60 rev

I chose 60 rev but calculated it wrong

A ball of mass m₁ moving at the velocity of +5m/s hits a second ball (which is stationary) of mass m₂. After the collision the velocity of m1 is found to be -2m/s. Select the correct answer from the followings (assume that the collision is elastic).

2m1=5m2

5m1=2m2

7m1=3m2

m1=m2

I chose A but got it wrong

please explain clearly with steps and formulas

Answer 1

1) For the spinning turbine, use the equation V_T = rω where VT is tangential velocity, r is the radius, and ω is the angular velocity. So:

V_T = rω

V_T = (2)(6) = 12 m/s (note that in the unit "radians/sec", the "radians" is basically just a placeholder and does not result in a unit when multiplied (meters x rad/sec = meters/sec not meter-radians/sec)

2) For the spinning compact disc, use ω_AVG = ½(ω_f + ω_i) which just basically says that the average angular speed is halfway between 30 and zero. So:

ω_AVG = ½(ω_f + ω_i)

ω_AVG = ½(0 + 30)

ω_AVG = 15 rev/min.

Now use θ = ωt

θ = ωt

θ = (15 rev/min)(2 min) = 30 rev

3) For the ball collision:

Momentum is conserved so P_i = P_f

P_i = m₁v_1i + m₂v_2i

P_i = m₁(5) + m₂(0) = 5m₁

P_f = m₁v_1f + m₂v_2f

P_f = m₁(-2) + m₂v_2f = -2m₁ + m₂v_2f

Since P_i = P_f

5m₁ = -2m₁ + m₂v_2f

7m₁ = m₂v_2f which can also be written as 7m₁/m₂ = v_2f

Also, since the collision is elastic, kinetic energy is also conserved so KE_i = KE_f

KE_i = 1/2m₁v_1i² + 1/2m₂v_2i²

KE_i = 1/2m₁(5)² + 1/2m₂(0)² = 25/2m₁

KE_f = 1/2m₁v_1f² + 1/2m₂v_2f²

KE_f = 1/2m₁(-2)² + 1/2m₂v_2f²

KE_f = 1/2(4)m₁ + 1/2m₂v_2f²

And since KE_i = KE_f then 25/2m₁ = 1/2(4)m₁ + 1/2m₂v_2f²

Multiplying both sides by 2 gives:

25m₁ = 4m₁ + m₂v_2f²

21m₁ = m₂v_2f² which can also be written as √(21m₁/m₂) = v_2f

Combining the equation from conservation of momentum and the equation from conservation of energy we get:

7m₁/m₂ = v_2f = √(21m₁/m₂)

7m₁/m₂ = √(21m₁/m₂) then squaring both sides gives:

49m₁²/m₂² = 21m₁/m₂

49m₁/m₂ = 21

49m₁ = 21m₂

7m₁ = 3m₂

Answer 2

Hi Ian C.,

I think I already gave you feedback on #3? Use conservation of momentum, and figure first the final momentum of ball m2: it is +7 m1 m/s . Note that that is in mass units involving m1, not m2! Next, calculate the initial and final kinetic energies of the total system, under each of the proposed m1/m2 ratios. All of these shuld be figured in terms of m1 based KE units! Only one of these will return KE(initial) exactly, that's your solution.

#1: A radian is a unit of measure for an angle (~57 degrees) such that the arc swung equals the leg (the radius of a partial circle). So, go 1 radian, you've gone the distance of the radius. Since your turbine blade is 2 m long, that's the radius of the circle that its tip rotates in. Then 6 rad/s * 2 m radius = 12 m/s

#2: assume that the CD slows unifrmly and linearly over time (you have to make some assumption of what it does, and this is a reasonable one, based on friction within its bearing. If you modeled as an exponential slowing process, then the CD would never completely stop, would it!). So, the average speed would be (1/2) the initial speed, or 15 rev/min . That, times 2 minutes = 30 rev .

Hope that helps you, -- Cheers, -- Mr. d.