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Problem 49

Question

A thin, uniform metal bar, 2.00 \(\mathrm{m}\) long and weighing \(90.0 \mathrm{N},\) is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 1.50 \(\mathrm{m}\) below the ceiling by a small 3.00 -kg ball, initially traveling horizontally at 10.0 \(\mathrm{m} / \mathrm{s} .\) The ball rebounds in the opposite direction with a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) . (a) Find the angular speed of the bar just after the collision. (b) During the collision, why is the angular momentum conserved but not the linear momentum?

Step-by-Step Solution

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Answer
(a) Angular speed \(\omega\) is found using \(\omega = \frac{1.50 \times \Delta p}{I}\). (b) Angular momentum is conserved because no external torques act on the bar-ball system around the pivot.
1Step 1: Identify Known Values
The problem states an initial velocity of the ball as \(v_i = 10.0 \, \text{m/s}\), a final velocity of the ball \(v_f = -6.0 \, \text{m/s}\), a mass of the ball \(m = 3.00 \, \text{kg}\), and the length of the bar as \(L = 2.00 \, \text{m}\). The bar pivots 1.50 m from the impact point.
2Step 2: Determine Moment of Inertia of the Bar
The moment of inertia \(I\) of a uniformly thin bar about an axis through one end is given by \(I = \frac{1}{3} mL^2\). The mass of the bar \(m_b\) is calculated from the weight as \(m_b = \frac{90.0 \, \text{N}}{9.8 \, \text{m/s}^2}\). Thus, \(I\) becomes \(I = \frac{1}{3} \times \frac{90.0}{9.8} \times (2.00)^2\).
3Step 3: Calculate Change in Linear Momentum of the Ball
The initial momentum of the ball is \(p_i = m \times v_i = 3.00 \times 10.0\). The final momentum is \(p_f = m \times v_f = 3.00 \times (-6.0)\). The change in momentum \(\Delta p = p_f - p_i = 3.00 \times (-6.0) - 3.00 \times 10.0\).
4Step 4: Use Conservation of Angular Momentum
The initial angular momentum of the ball is \(L_{i} = (1.50) \times p_i\). The final angular momentum is \(L_{f} = (1.50) \times p_f\). Conservation of angular momentum implies that \(L_{i} - L_{f} = I \times \omega\), where \(\omega\) is the angular speed.
5Step 5: Solve for Angular Speed \(\omega\)
Rearrange the equation for conservation of angular momentum to find \(\omega\). Substitute \(\Delta p = p_f - p_i\) and the calculated moment of inertia \(I\). Solve \(\omega = \frac{1.50 \times (\Delta p)}{I}\).
6Step 6: Explain Conservation of Angular Momentum
Linear momentum is not conserved because an external force acts on the system at the pivot. Angular momentum is conserved because the external force acts through the pivot point and does not exert a torque on the system.

Key Concepts

Angular SpeedMoment of InertiaLinear MomentumCollision Dynamics
Angular Speed
Angular speed refers to how fast an object rotates or spins around a central axis. In this context, it's the speed at which the bar starts rotating immediately after being struck by the ball. Angular speed is denoted by \(\omega\) and is measured in radians per second (rad/s).
Just like linear speed tells us how fast something moves in a straight line, angular speed reveals the rate of rotation. When an object like our bar rotates without changing its linear position, only the angular counterpart is significant.
  • To calculate angular speed, use the conservation of angular momentum.
  • The formula relates to angular speed through \( L = I \times \omega \), where \(L\) is angular momentum and \(I\) is moment of inertia.
Understanding angular speed is crucial for predicting how fast the bar will spin after the collision. Faster angular speeds mean faster spinning, just like a figure skater spins faster when they pull their arms in.
Moment of Inertia
moment of inertia acts as the "rotational mass" that describes how an object's mass is distributed around an axis, impacting how easily it starts rotating. In simple terms, it's tougher to spin an object with higher moment of inertia.
For a thin bar pivoting at one end, the moment of inertia is calculated using the formula:\[I = \frac{1}{3} m L^2\]where \(m\) is the mass of the bar and \(L\) is its length.
  • The higher the value of \(I\), the more resistance the object shows to changes in its rotation.
  • It plays a pivotal role in determining angular speed post-collision.
  • In our scenario, we use the bar's mass (derived from its weight) and length to compute its \(I\).
By having a clear understanding of moment of inertia, it's easier to ascertain why certain objects rotate faster than others when the same force is applied.
Linear Momentum
Linear momentum is all about an object's motion in a straight line. It's simply the product of an object’s mass and its velocity, expressed as \(p = m v\). In the scenario of the ball and bar, it represents how much motion the ball had before and after the collision.
During the collision, the ball's momentum changes because:
  • Initial momentum: \( p_i = m \times v_i = 3.00 \times 10.0 \)
  • Final momentum: \( p_f = m \times (-6.0) \)
This change in momentum, or \( \Delta p\), helps understand the forces involved. The ball's speed changes as it bounces back, indicating a shift in its linear momentum.

External Forces Influence