As shown in the figure, a bullet of mass m
and speed v passes completely through a
pendulum bob of mass M. The bullet
emerges with a speed of v/2. The
pendulum bob is suspended by a stiff rod
of length ℓ and negligible mass.
2.1 Derive an expression for the minimum value of v such that the pendulum bob
will barely swing through a complete vertical circle. (6)
2.2 Assuming the same value for v, obtain an expression for the centripetal
acceleration of the bob as it passes through the angular position +/2 radians
beyond its starting point.
momentum released to bob: mv/2
at impact, momentum
mv/2=(M )V so initial V is
V=m/(M ) v/2
so to make it to the top...
1/2 (M)((m/M)^2 v^2/4)=Mg L
then for min speed, solve for v
2.1 To derive the expression for the minimum value of v, we need to consider the conservation of energy in this system.
When the bullet passes through the pendulum bob, it transfers some of its kinetic energy to the bob. This kinetic energy is used to overcome the gravitational potential energy and provide the bob with enough velocity to complete a vertical circle.
Let's denote the velocity of the bullet before passing through the bob as v₁ and the velocity of the bullet after passing through the bob as v₂.
The initial kinetic energy of the bullet is (1/2)mv₁², and after passing through the bob, the kinetic energy of the bullet is (1/2)m(v₂)².
The energy transferred to the bob is given by (1/2)m(v₁)² - (1/2)m(v₂)².
This energy is used to raise the bob to its highest point, which corresponds to the potential energy of the bob at the highest point. The gravitational potential energy at the highest point is Mgℓ, where g is the acceleration due to gravity and ℓ is the length of the pendulum.
Setting the transferred energy equal to the gravitational potential energy, we have:
(1/2)m(v₁)² - (1/2)m(v₂)² = Mgℓ
Rearranging the equation, we get:
(1/2)m(v₁)² = (1/2)m(v₂)² + Mgℓ
Let's solve this equation for v₁:
(v₁)² = (v₂)² + 2gℓ
Taking the square root on both sides:
v₁ = √[(v₂)² + 2gℓ]
Now, to find the minimum value of v, we need to determine the minimum value of v₂. The minimum value of v₂ occurs when the bullet loses the maximum amount of energy, i.e., when v₂ = v/2.
Substituting v₂ = v/2 into the equation for v₁:
v₁ = √[(v/2)² + 2gℓ]
Thus, the expression for the minimum value of v is:
v_min = √[(v/2)² + 2gℓ]
2.2 To obtain the expression for the centripetal acceleration of the bob as it passes through the angular position +π/2 radians beyond its starting point, we need to consider the forces acting on the bob at that position.
At the angular position +π/2 radians beyond the starting point, the bob is at its highest point in the vertical circle. At this position, the net force acting on the bob is the centripetal force, provided by the tension in the rod.
The tension in the rod should be sufficient to provide the necessary centripetal force for circular motion. The centripetal force is given by:
F_c = Mw²ℓ
Where M is the mass of the bob, w is the angular velocity (w = v/ℓ), and ℓ is the length of the pendulum.
The centripetal acceleration can be determined using the formula:
a_c = w²ℓ
Substituting w = v/ℓ into the equation for centripetal acceleration:
a_c = (v/ℓ)²ℓ
Simplifying the equation:
a_c = v²/ℓ
Thus, the expression for the centripetal acceleration of the bob as it passes through the angular position +π/2 radians beyond its starting point is given by:
a_c = v²/ℓ
To solve this problem, we'll use conservation of momentum and energy principles.
2.1 Deriving the minimum value of v:
When the bullet passes through the pendulum bob, there is no external force acting on the system. Therefore, the total momentum before and after the collision should be the same.
Before the collision:
Momentum of the bullet = mv
Momentum of the pendulum bob = 0 (as it is at rest)
After the collision:
Momentum of the bullet = (m/2)(v/2) = mv/4 (as the bullet emerges with a speed of v/2)
Momentum of the pendulum bob = MV (as it starts moving)
Setting the momentum before and after the collision equal, we have:
mv = MV + mv/4
mv/4 = MV
v/4 = V
Now, we need to find the minimum value of v such that the pendulum bob barely completes a vertical circle. This means that the tension in the string should become zero at the topmost point of its circular motion. This happens when the centripetal force on the bob at the topmost point is equal to zero.
At the topmost point, the weight of the bob provides the centripetal force. So, we can equate the weight to the centripetal force:
mg = Mv^2/ℓ
Since we're looking for the minimum value of v, substitute V = v/4 into the equation:
mg = (M/16)v^2/ℓ
v^2 = (16mgℓ)/M
v = √((16mgℓ)/M)
Therefore, the expression for the minimum value of v is v = √((16mgℓ)/M).
2.2 Expressing the centripetal acceleration:
The centripetal acceleration of the bob at any angular position can be calculated using the formula:
a = v^2/ℓ
Since the angular position is +π/2 radians beyond its starting point, we substitute V = v/4 in the formula and calculate:
a = (v/4)^2/ℓ
a = (v^2/16)/ℓ
a = v^2/(16ℓ)
Therefore, the expression for the centripetal acceleration of the bob as it passes through the angular position +π/2 radians beyond its starting point is a = v^2/(16ℓ).