Answer:
Total angle through which the wheel has turned 58.9s after it begins rotating is 1709.52 rad
Explanation:
The image attached would offer a better explanation
Answer:
The answer to the question is;
The total angle through which the wheel has turned 58.9 s after it begins rotating is approximately 1709.67 rad.
Explanation:
To solve the question we note the equation for the motion of the flywheel as
ω₂ = ω₁ + α·t
Where:
ω₁ = Initial angular velocity = 0 rad/s as the body is initially at rest
ω₂ = Final angular velocity
α = angular acceleration = 1.35 rad/s²
t = Time = 28.3 s
Plugging in the values, we find ω₂
ω₂ = 0 + 1.35 rad/s²× 28.3 s = 38.205 rad/s
Since the acceleration is constant, only the mean velocity is required to determine the angle traveled during the first 28.3 seconds thus
Average velocity
ω=
= 19.1025 rad/s
The total angle traveled in 28.3 s is ω × time
= 19.1025 rad/s × 28.3 s = 540.60075 rad
After this the remaining time left is
58.9 s - 28.3 s = 30.6 s
Since the flywheel is moving at a constant velocity of 38.205 rad/s during the last 30.6 s we have
Angle traveled in 30.6 s at an angular velocity of 38.205 rad/s is given by
Angle traveled = Time × Angular velocity = 30.6 s × 38.205 rad/s
= 1169.073 rad
Therefore, the total angle traveled by the flywheel in 58.9 s is given by
540.60075 rad + 1169.073 rad = 1709.67375 rad ≈1709.67 rad.
the total angle traveled by the flywheel in 58.9 s ≈ 1709.67 rad.
Answer:
u= 20.09 m/s
Explanation:
Given that
m = 0.02 kg
M= 2 kg
h= 0.2 m
Lets take initial speed of bullet = u m/s
The final speed of the system will be zero.
From energy conservation
1/2 m u²+ 0 = 0+ (m+M) g h
m u²=2 (m+M) g h
By putting the values
0.02 x u² = 2 (0.02+2) x 10 x 0.2 ( take g=10 m/s²)
u= 20.09 m/s
