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:
a) v1 = 5.52m/s
b) v2 = -1.52m/s
c) v3 = 4.62m/s
d) vt = 3.85m/s
Explanation:
The velocity of the football wide receiver is his displacement per unit time.
Velocity v = (displacement d)/time t
v = d/t .....1
For each of the cases, equation 1 would be used to calculate the velocity.
a) v1 = d1/t1
d1= 16m
t1 = 2.9s
v1 = 16m/2.9s
v1 = 5.52m/s
b) v2 = d2/t2
d2 = -2.5m
t2 = 1.65s
v2 = -2.5/1.65
v2 = -1.52m/s
c) v3 = d3/t3
d3 = 24m
t3 = 5.2s
v3 = 24/5.2
v3 = 4.62m/s
d) vt = dt/tt
dt = 16m - 2.5m + 24m = 37.5m
tt = 2.9 + 1.65 + 5.2 = 9.75s
vt = 37.5/9.75
vt = 3.85m/s
Answer:
11.8 m/s
Explanation:
At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).
Sum of forces in the centripetal direction:
∑F = ma
mg − N = m v²/r
At the maximum speed, the normal force is 0.
mg = m v²/r
g = v²/r
v = √(gr)
v = √(9.8 m/s² × 14.2 m)
v = 11.8 m/s
Answer:
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