Question

A disk with mass m = 9.5 kg and radius r = 0.3 m begins at rest and accelerates uniformly for t = 18.1 s, to a final angular speed of ω = 28 rad/s. 1) what is the angular acceleration of the disk? rad/s2 2) what is the angular displacement over the 18.1 s? rad 3) what is the moment of inertia of the disk? kg-m2 4) what is the change in rotational energy of the disk? j 5) what is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s2 6) what is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed? m/s2 7) what is the final speed of a point on the disk half-way between the center of the disk and the rim? m/s 8) what is the total distance a point on the rim of the disk travels during the 18.1 seconds?

Answer 1

Ahhh this going to be confusing sorry...
1. α = Δω / Δt = 28 rad/s / 19s = 1.47 rad/s²

2. Θ = ½αt² = ½ * 1.47rad/s² * (19s)² = 266 rads

3. I = ½mr² = ½ * 8.7kg * (0.33m)² = 0.47 kg·m²

4. ΔEk = ½Iω² = ½ * 0.47kg·m² * (28rad/s)² = 186 J

5. a = α r = 1.47rad/s² * 0.33m = 0.49 m/s²

6. a = ω² r = (14rad/s)² * 0.33m = 65 m/s²

7. v = ω r = 28rad/s * ½(0.33m) = 4.62 m/s

8. s = Θ r = 266 rads * 0.33m = 88 m

Answer 2

1. The angular acceleration of the disk is = 1.55 rad/s² .
2. The angular displacement ( Θ )  of the mass over 18.1s =  253.90 rad.
3. The moment of inertia of the disk ( I ) = 0.43 kg.m²
4. The change in the rotational energy of the disk ( ΔEr ). = 168.56 J
5. The tangential component of the acceleration when disk have achieved half of final angular speed. = 0.465 m/s².
6. The magnitude of radial component when disk have achieved half of final angular speed  =  58.8 m/s².
7. The final speed of a point on the disk half-way between the center of disk and rim = 4.2 m/s
8. The total distance point in rim of the disk travels the entire time = 75.9m

Given data :

Mass ( m ) = 9.5 kg

Radius ( r ) = 0.3m

Time ( t ) = 18.1 s

Final angular speed ( ω ) = 28 rad/s

1) Calculate The angular acceleration of the disk suing the relation below

α = Δω / Δt

    = 28 rad/s / 18.1s ≈ 1.55 rad/s²

2) Calculating The angular displacement ( Θ )

 Θ = ½ *α*t²

    = ½ * 1.55rad/s² * (18.1)²  = 253.90 rad

3) Calculating the moment of inertia ( I )

I = ½*m*r²

  = ½ * 9.5kg * (0.3m)²

  = 0.43 kg.m²

4) Calculating the change in rotational energy of the disk ( ΔEr )

ΔEr = ½*I*ω²

       = ½ * 0.43kg·m² * (28rad/s)²

       = 168.56 J.

5) Calculating the tangential component of the acceleration

a = α*r

   = 1.55rad/s² * 0.3m

   = 0.465 m/s².

6) Determine the magnitude of radial component when disk have achieved half of final angular speed

 a = ω² r                           where ω = 28 rad / 2 =  14 rad

   = (14rad/s)² * 0.3m

   = 58.8 m/s².

7) Calculating the final speed of a point on the disk half-way between the center of disk and rim.

V = ω*r

   = 28rad/s * ½(0.3m)

    = 4.2 m/s.

8) Determine the total distance ( s ) a point in rim of the disk travels the entire time

 S = Θ* r

   = 253 rad * 0.3m

   = 75.9 m

Hence we can conclude that the answers to your question are as given above.

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