Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
Answer:
<u>Amplitude - remains the same</u>
<u>Frequency - increases</u>
<u>Period - decreases</u>
<u>Velocity - remains the same.</u>
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Explanation:
The amplitude of the wave remains the same since you are not changing the distance your hand moves and the amplitude of the wave depends on how much distance your hand covers while moving.
The frequency of your wave increases since now you are moving your hand more number of times in the same period i.e. your hand is moving faster in one second. So, the frequency of your wave increases.
The period is the time taken by the wave to travel a certain distance. Since your hand is now moving faster, the wave will travel faster and will take less time to cover the same distance hence, we can say that its period will decrease.
The velocity of a wave depends on the medium in which it is travelling. Your wave was previously travelling in air and the new wave is also travelling in the same medium so the velocity of the wave remains unchanged.
