Answer: Thus, the force is directed 72.5° above the horizontal.
Explanation:
The magnitude of the force, F = 15 N
let this force be detected at angle θ from the horizontal, then horizontal component of force is: F cos θ
and vertical component is F sin θ
Horizontal component is given,
F cos θ = 4.5 N
⇒15 N cos θ = 4.5 N
⇒ cos θ = 0.3
⇒ θ = cos⁻¹ 0.3 = 72.5°
The uncharged body will gain electrons from the charged body. Giving the uncharged body a shock.
The moment the stick comes to rest at θ=62.1° from horizontal.
<span>Angular acceleration = (net torque) / (moment of inertia) </span>
<span>α = τ/I </span>
<span>We have to add up the torques due to the bugs and the stick; and add up the moments of inertia due to all three also. </span>
<span>Let L be the stick's length and let m be the stick's mass (so "2.75m" is each bug's mass). And let's say the "lower" ladybug is on the left. Then the lower ladybug exerts this much torque: </span>
<span>τ_lowerbug = −(2/5)L(2.75mg)cosθ (negative because I am (arbitrarily) choosing counter-clockwise as the negative angular direction). </span>
<span>The upper ladybug exerts this much torque: </span>
<span>τ_upperbug = +(3/5)L(2.75mg)cosθ </span>
<span>The weight of the stick can be assumed to act through its center, which is 1/10 of the way from the fulcrum. So the stick exerts this much torque: </span>
<span>τ_stick = +(1/10)L(mg)cosθ </span>
<span>The net torque is thus: </span>
<span>τ_net = τ_lowerbug + τ_upperbug + τ_stick </span>
<span>= −(2/5)L(2.75mg)cosθ + (3/5)L(2.75mg)cosθ + (1/10)L(mg)cosθ </span>
<span>= (2.75(3/5−2/5)+1/10)(mgL)cosθ </span>
<span>Now for the moments of inertia. The bugs can be considered point masses of "2.75m" each. So for each of them you can use the simple formula: I=mass×R²: </span>
<span>I_lowerbug = (2.75m)((2/5)L)² = (2.75m)(4/25)L² </span>
<span>I_upperbug = (2.75m)((3/5)L)² = (2.75m)(9/25)L² </span>
<span>For the stick, we can use the parallel axis theorem. This says, when rotating something about an axis offset a distance "R" from its center of mass, the moment of inertia is: </span>
<span>I = I_cm + mR² </span>
<span>We know that for a stick about its center of mass, I_cm is (1/12)mL² (see many sources). And in this problem we know that it's offset by R=(1/10)L. So: </span>
<span>I_stick = (1/12)mL² + m((1/10)L)² </span>
<span>= (1/12)mL² + (1/100)mL² </span>
<span>= (7/75)mL² </span>
<span>So the total moment of inertia is: </span>
<span>I_total = I_lowerbug + I_upperbug + I_stick </span>
<span>= (2.75m)(4/25)L² + (2.75m)(9/25)L² + (7/75)mL² </span>
<span>= (2.75(4/25+9/25)+7/75)mL² </span>
<span>So that means the angular acceleration is: </span>
<span>α = τ_net/I_total </span>
<span>= ((2.75(3/5−2/5)+1/10)(mgL)cosθ)/((2.75(4... </span>
<span>The "m" cancels out. You're given "L" and "θ" and you know "g", so do the math (and don't forget to use consistent units).</span>
The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s