A rod is lying at rest on a perfectly smooth horizontal surface (no friction). We give rod a short impulse (a hit) perpendicular to the length direction of the rod at X. The mass of the rod is 3 kg, and its length is 50 cm. The impulse is 4 kg⋅m/sec. The distance from the center C of the rod to X is 15 cm.
(a) What is the translational speed ∣∣v⃗ cm∣∣ of C after the rod is hit? (in m/s)
∣∣v⃗ cm∣∣=
(b) What is the magnitude of the angular velocity ω of the rod about C? (in radians/s)
ω=
(c) How far (distance D in meters) has the center C of the rod moved from its initial position 8 seconds after it was hit? D=
And what is the angle θ (in radians) between the direction of the rod at 8 seconds after it was hit, and its initial direction (before it was hit)? Give the smaller angle.
θ(smaller angle)=
(d) What is the total kinetic energy K of the rod after it was hit? (in Joules)
K=
Problem 7.9
http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/assignments/sol7.pdf
You didn't take a look at the help sessions, did you? Watch the 6th video of HS. It is 95% done there.
To solve this problem, we can use the principle of conservation of angular momentum and the equation of conservation of linear momentum.
(a) To find the translational speed of the center of mass (C) after the rod is hit, we can use the conservation of linear momentum.
The initial linear momentum of the rod is zero since it is at rest. The impulse applied to the rod is given as 4 kg⋅m/sec. The mass of the rod is 3 kg.
Impulse = Change in momentum
Impulse = mass × change in velocity
4 kg⋅m/sec = 3 kg × ∣∣v⃗ cm∣∣
∣∣v⃗ cm∣∣ = 4 kg⋅m/sec / 3 kg
∣∣v⃗ cm∣∣ ≈ 1.33 m/s
So, the translational speed of the center of mass (C) after the rod is hit is approximately 1.33 m/s.
(b) To find the magnitude of the angular velocity of the rod about C, we can use the conservation of angular momentum.
The initial angular momentum of the rod is zero since it is at rest. The impulse applied to the rod creates an angular momentum.
Change in angular momentum = Impulse × distance from the center
Change in angular momentum = 4 kg⋅m/sec × 0.15 m (since the distance from the center C to X is 15 cm)
Change in angular momentum = 0.6 kg⋅m²/sec
The moment of inertia (I) of a rod rotating about its center is given by I = (1/12) × mass × length².
I = (1/12) × 3 kg × (0.5 m)²
I = 0.625 kg⋅m²
Angular momentum = Moment of inertia × angular velocity
0.6 kg⋅m²/sec = 0.625 kg⋅m² × ω
ω = 0.6 kg⋅m²/sec / 0.625 kg⋅m²
ω ≈ 0.96 radians/sec
So, the magnitude of the angular velocity of the rod about C is approximately 0.96 radians/sec.
(c) To find the distance that the center C of the rod has moved from its initial position after 8 seconds, we can use the equation of motion for a uniformly accelerated object.
The initial velocity of C is approximately 1.33 m/s from part (a). The acceleration of C is zero since there is no external force acting on it.
Distance (D) = Initial velocity × time
D = 1.33 m/s × 8 sec
D ≈ 10.64 meters
So, the center C of the rod has moved approximately 10.64 meters from its initial position after 8 seconds.
To find the angle θ between the direction of the rod at 8 seconds after it was hit and its initial direction, we can use the tangent of the angle.
θ = arctan(D / (0.5 × length))
θ = arctan(10.64 m / (0.5 × 0.5 m))
θ ≈ 84.3 degrees
However, we need to give the angle in radians, so we convert 84.3 degrees to radians:
θ (smaller angle) = 84.3 degrees × (π / 180)
θ (smaller angle) ≈ 1.47 radians
So, the angle θ (smaller angle) between the direction of the rod at 8 seconds after it was hit and its initial direction is approximately 1.47 radians.
(d) To find the total kinetic energy of the rod after it was hit, we can use the equation for rotational kinetic energy and translational kinetic energy.
The translational kinetic energy (Kt) is given by Kt = (1/2) × mass × velocity².
Kt = (1/2) × 3 kg × (1.33 m/s)²
Kt ≈ 2.98 Joules
The rotational kinetic energy (Kr) is given by Kr = (1/2) × moment of inertia × angular velocity².
Kr = (1/2) × 0.625 kg⋅m² × (0.96 radians/sec)²
Kr ≈ 0.28 Joules
Total kinetic energy (K) = Kt + Kr
K ≈ 2.98 Joules + 0.28 Joules
K ≈ 3.26 Joules
So, the total kinetic energy of the rod after it was hit is approximately 3.26 Joules.