Answer:
Time, t = 2 seconds
Explanation:
Given the following data;
Mass, m = 50 kg
Initial velocity, u = 0 m/s (since it's starting from rest).
Final velocity, v = 8 m/s
Force, F = 200 N
To find the time, we would use the following formula;
Making time, t the subject of formula, we have;
Substituting into the formula, we have;
Time, t = 2 seconds
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
to find
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:
Answer:
Explanation:
Component of force perpendicular to stick
= F Sin 60°
=√3 / 2 F.
Taking torque about the other end
= √3 / 2 F x 1 Nm
Weight of stick = 60 gm
= 60 x 10⁻³ kg
= 60 x 10⁻³ x 9.8 N
= .588 N
This weight will act from the middle point of stick so torque about the
other end
= .588 x 1 Nm
Balancing these two torques we have
.588 = √3 /2 F
F = 0.679 N