A crate is placed on an adjustable, incline board. the coefficient of static friction between the crate and the board is 0.29.
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Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis
KE = 0.23 J
given parameters
- Disk radius R = 15 cm = 0.15 m
- Cylinder radius r = 1.5 cm = 0.0015 m
- Disk mass M = 20 kg
- Time t = 1.2 s
- Force F = 12 N
to find
- Kinetic energy (KE)
This exercise must be solved in parts:
1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation
KE = ½ I w²
Where KE is the kinetic energy, I the moment of inertia and w the angular velocity
The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is
I_{ total} = I_{ disk} + I_{ cylinder}
the moments of inertia with respect to an axis passing through the center of mass are tabulated
disk I_{disk} = ½ M R²
cylinder I_{cylinder} = ½ m r²
where M and m are the masses of the disk and cylinder respectively, R and r their radii
I_{total} = ½ (M R² + m r²) = ½ M R² ( )
I_{total} = ½ M R² ( ) =
M R² (1 + 0.005 m)
As the shaft mass is much lighter than the disk mass , the last term is very small, which is why we despise it.
I_{total} = ½ M R²
2nd part. Let's use Newton's second law for endowment motion
τ = I α
α = l
τ = F R
α =
With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)
w = w₀ + α t
where w is the angular velocity, alpha is the angular acceleration and t is the time
w = 0 +
we substitute in the kinetic energy equation
KE = ½ I_{total} ( )²
KE = ½
let's substitute
KE =
KE = F² R² t² / M
let's calculate
KE = 12² 0.15² 1.2² / 20
KE = 0.23 J
With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is
KE = 0.23 j
learn more about rotational kinetic energy here: