Let's break down the problem step by step: ### 1. **Given Data** - Mass of rod, \( M = 8.9 \) kg - Mass of bullet, \( m = 5.9 \) g = \( 0.0059 \) kg - Bullet speed, \( v = 640 \) m/s - Impact point from the axis, \( d = 0.15 \) m - Gravity, \( g = 9.81 \) m/s\(^2\) - The rod is hung from a rope (assume the axis is at one end). ### 2. **Find the Velocity after Collision** - The collision is inelastic (bullet embeds itself). - **Angular momentum** is conserved about the pivot point (top of the rod). Initial angular momentum (about the pivot): \[ L_{i} = m v d \] After collision, bullet and rod rotate together: \[ L_{f} = I_{\text{total}} \omega \] Where: - \( I_{\text{rod,pivot}} = \frac{1}{3} M L^2 \) - \( I_{\text{bullet,pivot}} = m d^2 \) But **rod length \( L \) is not given**. But for finding the instantaneous center of rotation (IC), you do **not** need \( L \) if the answer is in terms of \( d \). ### 3. **Find the Angular Velocity after Collision** Set angular momentum before = after: \[ m v d = \left( \frac{1}{3} M L^2 + m d^2 \right)\omega \] \[ \omega = \frac{m v d}{\frac{1}{3} M L^2 + m d^2} \] But to find the IC, we need the **translational velocity** of the center of mass and the **angular velocity**. ### 4. **Find the Velocity of the Center of Mass after Collision** After collision, the **center of mass** of the rod-bullet system has a velocity perpendicular to the rod (since rod starts from rest): \[ v_{CM} = \omega r_{CM} \] Where \( r_{CM} \) is the distance from the pivot to the new center of mass: \[ r_{CM} = \frac{M \frac{L}{2} + m d}{M + m} \] But since the bullet is much lighter than the rod, \( r_{CM} \approx \frac{L}{2} \). ### 5. **Instantaneous Center of Rotation (ICR) Calculation** The ICR is the point about which the rod *appears* to rotate at the instant after the collision. For a rigid body: \[ v_{CM} = \omega h \] where \( h \) is the **distance from the center of mass to the ICR**. But, more generally, **distance from the pivot**: \[ h = \frac{v_{CM}}{\omega} \] But for a rod rotating about a fixed point, the ICR is **at the pivot**. However, if the pivot is not fixed (e.g., the rope is slack or the rod is momentarily in motion), then the ICR is below the pivot. #### For a rod swinging about a pivot, the IC is at a point below the pivot along the rod, such that the velocity of the center of mass is perpendicular to the rod and equal to \(\omega \frac{L}{2}\). #### The velocity of the end (pivot) is zero (because it's hung by the rope). #### The velocity at any point a distance \( h \) below the pivot is: \[ v = \omega h \] So for the **center of mass**: \[ v_{CM} = \omega \cdot \frac{L}{2} \] So, the ICR is a distance \( h \) above the center of mass where the velocity is zero. The distance from the pivot to the ICR is: \[ h_{ICR} = \frac{L}{2} \] But sometimes the question wants the vertical distance below the pivot, which is the same as the length from pivot to center of mass: **for a uniform rod, this is \( \frac{L}{2} \) below the pivot.** ### 6. **But your question gives numbers for the mass and the bullet impact point, but NOT the rod length.** #### Let's see if we need to use the bullet info. Actually, for a rod pivoted at one end, the **instantaneous center of rotation right after an impulse at a point \( d \) from the pivot** is at a distance: \[ h = \frac{I}{M r_{CM}} \] for a uniform rod, \( I = \frac{1}{3}ML^2 \), \( r_{CM} = \frac{L}{2} \): \[ h = \frac{\frac{1}{3}ML^2}{M \frac{L}{2}} = \frac{2}{3}L \] So the answer is \( h = \frac{2}{3}L \) **below the pivot**. But, since you are given the impact point \( d \), and usually the question expects you to use that, perhaps the actual answer is: \[ h = \frac{I}{M d} \] with \( d \) as the impact point. So, \[ h = \frac{\frac{1}{3} M L^2}{M d} = \frac{L^2}{3d} \] But this doesn't have units of length unless \( L \) is given. ### **BUT**: In most problems of this type, the IC is at a distance \( h = \frac{L}{2} \) from the pivot for a uniform rod, regardless of the impact (assuming the rod is fixed at the pivot). If the pivot is not fixed (e.g., if it's suspended by a rope), the IC might be lower, depending on the impulse. #### Given the data, the only possible answer (without \( L \)) is the **distance from the pivot to the point of impact**: Therefore, the **IC** is at the **pivot point** (i.e., at the top of the rod). ### **Summary** #### If the rod is hung from a rope and pivots about the top, the instantaneous center of rotation (IC) is **at the pivot**, i.e., \( h = 0 \) m. #### But if the question is asking for the distance from the center of mass to the IC, it is \( \frac{L}{2} \). #### If the question is asking for the distance from the pivot to the IC **immediately after the bullet embeds** at distance \( d \), **the answer is**: \[ h = \frac{I}{M d} \] where \( I = \frac{1}{3} M L^2 \), so \[ h = \frac{L^2}{3d} \] But since \( L \) is not given, **the only numerical answer you can give is**: \[ \boxed{0.00 \text{ m}} \] **If you are required to use the numbers and the formula, or if the question wants the answer in terms of \( d \), plug in \( d = 0.15 \) m:** If you use: \[ h = \frac{I}{M d} = \frac{\frac{1}{3} M L^2}{M d} = \frac{L^2}{3d} \] But still, \( L \) is not given. --- ## **Final Answer** Given the information provided and the standard physics of this problem, **the instantaneous center of rotation is at the pivot point, thus:** \[ \boxed{h = 0.00 \text{ m}} \] **If you are required to enter a nonzero answer, clarify if \( L \) is given or expected. If not, \( 0.00 \) m is the correct answer.**