Answer:
(A) 0.63 J
(B) 0.15 m
Explanation:
length (L) = 0.75 m
mass (m) =0.42 kg
angular speed (ω) = 4 rad/s
To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)
I = Ic + m
Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis
h is the horizontal distance between the center of mass and the rotational axis of the rod
I = ![(\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B12%7D%29%28mL%5E%7B2%7D%20%29%20%2B%20m%28%5Btex%5D%5Cfrac%7BL%7D%7B2%7D)
)^{2}[/tex]
I = ![(\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B12%7D%29%280.42%20x%200.75%5E%7B2%7D%20%29%20%2B%20%28%200.42%20x%20%28%5Btex%5D%5Cfrac%7B0.75%7D%7B2%7D)
)^{2}[/tex])
I = 0.07875 kg.m^{2}
(A) rods kinetic energy = 0.5I
= 0.5 x 0.07875 x 
= 0.63 J 0.15 m
(B) from the conservation of energy
initial kinetic energy + initial potential energy = final kinetic energy + final potential energy
Ki + Ui = Kf + Uf
at the maximum height velocity = 0 therefore final kinetic energy = 0
Ki + Ui = Uf
Ki = Uf - Ui
Ki = mg(H-h)
where (H-h) = rise in the center of mass
0.63 = 0.42 x 9.8 x (H-h)
(H-h) = 0.15 m
“Don't hand that holier than thou line to me” is what the asymptote
said to the removable discontinuity.
The distance between the
curve and the line where it approaches zero as they tend to infinity is the line in the asymptote
of a curve. This is unusual for modern authors but in some
sources the requirement that the curve may not cross the line infinitely often
is included.
The point that does not fit the rest of the graph or is
undefined is called a removable discontinuity. By filling in a single
point, the removable discontinuity can be made connected.
Answer: D. ➡️⬅️
Explanation: I just knew the answer ;)
Answer:
Explanation:
Newton's 2nd law is given as 
.
To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.
Using trigonometry to find the horizontal component of the force:
Use this horizontal component of the force to solve for for the acceleration of the object:
The complete question was calculate the period T assuming the smallest amplitude.
Using the equation;
T = 2 π√(L/g)
Where T is the period in seconds, L is the length of the rod or wire in meters and g is the acceleration due to gravity.
Hence; T = 2×3.14 × √(2/9.81)
= 6.28 × 0.4515
= 2.836 seconds
