Question

ou are making a clock out of a solid disk with a radius of 0.2m and mass of 0.1 kg that will be attached at its center of mass to the end of a uniform thin rod with a mass of 0.3 kg. Calculate The length of the rod such that the period of the system is 1 second.The distance from the pivot to theCOM of the system must be expressed in terms of L. The parallel axis theorem must be used for both therod and the solid disk in terms of L as well. There will be a quadratic equation to be solved

Answer 1

the correct result is L = 0.319 m

Step-by-step explanation:

This system is a physical pendulum whose angular velocity is

w² =
`(M \ g \ d)/(I)`

where d is the distance from the center of mass to the point of rotation and I is the moment of inertia of the system

The Moment of Inertia is a scalar, therefore an additive quantity

I = I_bar + I_disk

the moment of inertia of each element with respect to the pivot point can be found with the parallel axes theorem

let's use M for the mass of the bar and m for the mass of the disk

Bar

I_bar = I_{cm} + Md²

the moment of inertia of the center of mass is

I_{cm} =
`(1)/(12)` M L²

the distance from the center of mass

d = L / 2

we substitute

I_bar =
`(1)/(12)` M L² + M (
`(L^2)/(4)`)

Disk

I_disk = I_{cm} + m d²

moment of inertia of the center of mass

I_{cm} = ½ m R²

the distance d is

d = L

we substitute

I_disk = 1/2 m R² + m L²

the total moment of inertia is

I =
`(1)/(12)` M L² +
`(1)/(4)` M L² +
`(1)/(2)` m r² + m L²

I =
`(1)/(4)` M L² + m L² + ½ m r²

I = L² (m +
`(1)/(4)` M) + ½ m r²

The position of the center of mass of the system can be found with the expressions

d_{cm} =
`(1)/(M) \sum r_i m_i`

d_{cm} =
`(1)/(m+M) \ ( M (L)/(2) + m L)`

d_{cm} =
`L (m + M/2)/(m +M )`

now we can substitute in the expression for the angular velocity

w² = (m + M) g L
`(m + (M)/(2) )/(m+M)`
`(1)/(L^2 (m+ (M)/(4) ) + (1)/(2) m r^2 )`

w² = g (m +
`(1)/(2)` M)
`(L)/( L^2 ( m +(1)/(4) M ) + (1)/(2) m r^2)`

angular velocity and period are related

w = 2π/T

sustitute

4π²/T² = g (m +
`(1)/(2)` M)
`(L)/( L^2 ( m +(1)/(4) M ) + (1)/(2) m r^2)`

L² (m +
`(1)/(4)` M) + ½ m r² =
`(T^2)/(4 \pi ^2 ) \ g ( m + (1)/(2) M ) \ \ L`

we substitute the values ​​and solve the second grade equation

L² (0.1 +
`(1)/(4)` 0.3) - [
`(1^2)/(4\pi ^2)` 9.8 (0.1 + 0.3/2) ] L + ½ 0.1 0.2² = 0

L² 0.175 - 0.06206 L + 0.002 = 0

the equation remains after simplifying

L² - 0.3546 La + 0.01143 = 0

solve us

L =
`(0.3546 \ \pm √( 0.3546^2 - 4 \ 0.01143 ))/(2)`

L =
`(0.3546 \ \pm \ 0.28288 )/(2)`

L₁ = 0.319 m

L₂ = 0.036m

the correct result must have a value greater than the radius of the disk. The correct result is L = 0.319 m