Relating the Speed of a Falling Block to Rotating Disks

[Science Student]

2 metal disks are welded together. One has a radius of R₁ = 0.025m with a mass of m₁ = 0.80kg and the other has a radius of R₂ = 0.05m with a mass of m₂ = 1.60kg. A light string is wrapped around the edge of the smaller disk. At the end of the string is a block with mass m_b = 1.50kg. The block is released y = 2.00m from the ground. What is the speed of the block just before it strikes the ground?

 

U(1) + K(1) = U(2) + K(2

 

U(1) + 0 = 0 + K(2)

 

U(1) = m_bgy

 

K(2) = (1/2)m_b(v^2) + (1/2)(I)(w^2),

 

where I = (1/2)(m₁) (R₁)^2 + (1/2)(m₂) (R₂)^2 = 0.0025kgm^2,

 

and w^2 = (v^2)/(R₁^2).

 

m_bgy = (1/2)m_b(v^2) + (1/2)(I)(w^2)

 

2m_bgy = m_b(v^2) + (I)(v^2)/(R₁^2)

 

v^2 = (2m_bgy)/(m_b + (I/(R₁^2)))

 

v = 3.27m/s

 

But the answer in the textbook is 3.40m/s. The textbook seems to never be wrong. Does anyone see anything wrong with my work?

 

 

[swansont]

I get 0.00225 for the moment of inertia calculation

[Science Student]

I get 0.00225 for the moment of inertia calculation

Oh thank-you so much!!!! I checked my work about 15 times and could not see that.