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 =
)^{2}[/tex]
I =
)^{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
Answer:
Part a)

Part b)

Explanation:
Part A)
As we know that time period of the motion is given as

so we have



now at the point of maximum amplitude the force equation when Normal force is about to zero is given as

so we have



Part b)
Now if the amplitude of the SHM is 6.23 cm
and now at this amplitude if object will lose the contact then in that case again we have



so now we have


C because mass is how much space an object takes up
Answer:
41.3 m/s^2 option (e)
Explanation:
force, F = 6.81 N
mass, m = 165 g = 0.165 kg
Let a be the acceleration of the puck.
Use newtons' second law
Force = mass x acceleration
6.81 = 0.165 x a
a = 41.27 m/s^2
a = 41.3 m/s^2
Thus, the acceleration of the puck is 41.3 m/s^2.