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
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➷ Earth's gravity is approximately 9.81
weight = mass x gravity
weight = 4.6 x 9.81
weight = 45.126
Answer is B. 45N
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Answer:
<em>600N.</em>
Explanation:
From the question, we are to calculate the net force acting on the car.
According to Newton's second law of motion:
F = ma
m is the mass of the car
a is the acceleration = change in velocity/Time
a = v-u/t
F = m(v-u)/t
v is the final velocity = 30m/s
u is the initial velocity = 20m/s
t is the time = 5secs
m = 300kg
Get the net force:
Recall that: F = m(v-u)/t
F = 300(30-20)/5
F = 60(30-20)
F = 60(10)
<em>F = 600N</em>
<em>Hence the net force acting on the car is 600N.</em>
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