Answer:
N_c = 3.03 N
F = 1.81 N
Explanation:
Given:
- The attachment missing from the question is given:
- The given expressions for the radial and θ direction of motion:
r = 0.5*θ
θ = 0.5*t^2 ...... (correction for the question)
- Mass of peg m = 0.5 kg
Find:
a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.
b) Determine the magnitude of the normal force of the slot on the peg.
Solution:
- Determine the expressions for radial kinematics:
dr/dt = 0.5*dθ/dt
d^2r/dt^2 = 0.5*d^2θ/dt^2
- Similarly the expressions for θ direction kinematics:
dθ/dt = t
d^2θ/dt^2 = 1
- Evaluate each at time t = 2 s.
θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°
r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2
- Evaluate the angle ψ between radial and horizontal direction:
tan Ψ = r / (dr/dθ) = 1 / 0.5
Ψ = 63.43°
- Develop a free body diagram (attached) and the compute the radial and θ acceleration:
a_r = d^2r / dt^2 - r * dθ/dt
a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2
a_θ = r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)
a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2
- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:
Radial direction: N_c * cos(26.57) - W*cos(24.59) = m*a_r
θ direction: F - N_c * sin(26.57) + W*sin(24.59) = m*a_θ
Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:
N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5
N_c = 3.03 N
F - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5
F = 1.81 N
A) The magnitude of the force of the rod on the peg at the instant t = 2 s is; F = 4.21 N
B) The magnitude of the normal force of the slot on the peg is; N = 4.90 N
We are given;
r = 0.5 θ
dr/dθ = 0.5
θ = (π/8)t²
Now The angle ψ between the extended radial line and the tangent can be gotten from;
tan ψ = r/(dr/dθ)
tan ψ = 0.5 θ/(0.5)
tan ψ = θ
A) At t = 2 s, θ = (π/8)2² = π/2 rad
Since tan ψ = θ, then we have;
tan ψ = π/2
Thus; ψ = tan⁻¹(π/2)
ψ = 57.52°
From the image of the peg attached, the free body diagram of the peg will mean that;
ΣF_r = ma_r; Nsinψ - mg = ma_r
Nsin57.2° - (0.5*9.81) = 0.5a_r ---(eq 1)
ΣF_θ = ma_θ; F - Ncos ψ = ma_θ
F - Ncos 57.2° = 0.5a_θ ---(eq 2)
Since, r = 0.5 θ and θ = (π/8)t², let us use chain rule to find the first and second derivatives of r and θ with respect to t. Thus;
r = 0.5((π/8)t²
r = ¹/₁₆πt²
dr/dt = ¹/₈πt
d²r/dt = ¹/₈π
Similarly;
θ = (π/8)t²
dθ/dt = ¹/₄πt
d²θ/dt = ¹/₄π
When t = 2;
r = ¹/₁₆π(2²) = ¹/₄π m
dr/dt = ¹/₈π(2) = ¹/₄π m/s
d²r/dt = ¹/₈π m/s²
θ = (π/8)(2²) = ¹/₂π rad
dθ/dt = ¹/₄π(2) = ¹/₂π rad/s
d²θ/dt = ¹/₄π rad/s²
Thus;
a_r = d²r/dt - r(dθ/dt)²
a_r = ¹/₈π - ¹/₄π(¹/₂π)²
a_r = -1.5452 m/s²
a_θ = r(d²θ/dt) + 2r(dr/dt)(dθ/dt )
a_θ = ¹/₄π(¹/₄π) + 2(¹/₄π)(¹/₄π)(¹/₂π)
a_θ = 3.0843 m/s²
Plugging in the relevant values into eq 1 and eq 2 gives;
Nsin57.2° - (0.5*9.81) = 0.5(-1.5452) ---(eq 1a)
F - Ncos 57.2° = 0.5(3.0843) ---(eq 2a)
⇒
0.8406N - 4.905 = -0.7726
N = 4.92 N
F - Ncos 57.2° = 0.5(3.0843)
F - 0.5417*4.92 = 1.5422
F = 4.21 N
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