A uniform thin rod of length 0.400 m and mass 4.40 kg can rotate in a horizontal plane about a vertical axis through its center.
the rod is at rest when a 3.0 g bullet traveling in the rotation plane is fired into one end of the rod. as viewed from above, the bullet's path makes angle θ = 60° with the rod. if the bullet lodges in the rod and the angular velocity of the rod is 17 rad/s immediately after the collision, what is the bullet's speed just before impact?
1 answer:
The rod has a mass of m = 4.4 kg and a length of L = 0.4 m.
Its polar moment of inertia is
J = (mL²)/12
= (1/12) * [(4.4 kg)*(0.4 m)²]
= 0.05867 kg-m²
The mass of the bullet is 0.3 g.
If its velocity is v m/s, then its linear momentum is
P = (0.3 x 10⁻³ kg)*(v m/s)
Its linear momentum perpendicular to the rod is
P*sin(60°) = 2.5981 x 10⁻⁴ v (kg-m)/s
The angular momentum about the center of the rod when the bullet strikes is
T = (2.5981 x 10⁻⁴ v (kg-m)/s)*(0.2 m) = 5.1962 x 10⁻⁵ v (kg-m²)/s
Because the bullet lodges into the end of the rod, the combined polar moment of inertia is
J + (0.3 x 10⁻³ kg)*(0.2 m)² = 0.05867 + 1.2 x 10⁻⁵ = 0.0587 kg-m²
The initial angular velocity is ω = 17 rad/s.
Because angular momentum is conserved, therefore
5.1962 x 10⁻⁵ v (kg-m²)/s = (0.0587 kg-m²)*(17 rad/s)
v = 19204 m/s
Answer: 19204 m/s
Its polar moment of inertia is
J = (mL²)/12
= (1/12) * [(4.4 kg)*(0.4 m)²]
= 0.05867 kg-m²
The mass of the bullet is 0.3 g.
If its velocity is v m/s, then its linear momentum is
P = (0.3 x 10⁻³ kg)*(v m/s)
Its linear momentum perpendicular to the rod is
P*sin(60°) = 2.5981 x 10⁻⁴ v (kg-m)/s
The angular momentum about the center of the rod when the bullet strikes is
T = (2.5981 x 10⁻⁴ v (kg-m)/s)*(0.2 m) = 5.1962 x 10⁻⁵ v (kg-m²)/s
Because the bullet lodges into the end of the rod, the combined polar moment of inertia is
J + (0.3 x 10⁻³ kg)*(0.2 m)² = 0.05867 + 1.2 x 10⁻⁵ = 0.0587 kg-m²
The initial angular velocity is ω = 17 rad/s.
Because angular momentum is conserved, therefore
5.1962 x 10⁻⁵ v (kg-m²)/s = (0.0587 kg-m²)*(17 rad/s)
v = 19204 m/s
Answer: 19204 m/s
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Answer:
Total contraction on the Bar = 1.22786 mm
Explanation:
Given that:
Total Length for aluminum bar = 600 mm
Diameter for aluminum bar = 40 mm
Hole diameter = 30 mm
Hole length = 100 mm
elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²
compressive load P = 180 KN = 180 × 10³ N
Calculate the total contraction on the bar = ???
The relation used in calculating the contraction on the bar is:
The relation used in calculating the total contraction on the bar can be expressed as :
Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)
i.e
Total contraction on the Bar =
Let's find the area of cross section without the hole and with the hole
Area of cross section without the hole is :
Using A = πd²/4
A = π (40)²/4
A = 1256.64 mm²
Area of cross section with the hole is :
A = π (40²-30²)/4
A = 549.78 mm²
Total contraction on the Bar =
Total contraction on the Bar =
Total contraction on the Bar = 2.117( 0.398 + 0.182)
Total contraction on the Bar = 2.117*(0.58)
Total contraction on the Bar = 1.22786 mm
Answer:
option (B)
Explanation:
Intensity of unpolarised light, I = 25 W/m^2
When it passes from first polarisr, the intensity of light becomes
Let the intensity of light as it passes from second polariser is I''.
According to the law of Malus
Where, θ be the angle between the axis first polariser and the second polariser.
I'' = 11.66 W/m^2
I'' = 11.7 W/m^2