IV. Off Center Bullet/Block Calculations Recall what happened when you looked at the video of an identical block rising

[answerhappygod]

Iv Off Center Bullet Block Calculations Recall What Happened When You Looked At The Video Of An Identical Block Rising 1 (77.12 KiB) Viewed 23 times

Iv Off Center Bullet Block Calculations Recall What Happened When You Looked At The Video Of An Identical Block Rising 2 (93 KiB) Viewed 23 times

IV. Off Center Bullet/Block Calculations Recall what happened when you looked at the video of an identical block rising just after being hit off-center. Even though the block was spinning rapidly, it rose to about the same height as the block that was hit "on-center!" This could be explained if the bullet didn't penetrate as far so that more of its energy was left to cause the block to rise and spin rapidly. The idea that the off-center bullet doesn't penetrate as far makes sense for two reasons: (1) If the block spins away from the bullet then the bullet won't have as much time to penetrate into it and (2) since you just showed that in an on-center shor almost 99% percent of the bullet's energy is lost plowing into the block, perhaps the off-center bullet doesn't penetrate quite as deep and there's enough energy to start the block spinning rapidly. NOTE: We expect variations in bullet speeds and block densities to cause differences in block heights from shot to shot, even when other factors are the same. What surprises most observers is that on average, blocks shot on-center and off-center both seems to rise to about the same height. Fig. 7: Photos of just the top of each path for eight blocks shot into the air on center" under "identical conditions." These height variations are probably due to differences in the block densities and each bullet's muzzle velocity Is the rotational kinetic energy of a spinning block hit "off-center” significant compared to a block's kinetic energy when shot on-center? To see if this rotational kinetic energy is significant, you can perform sample rotational energy calculations for identical blocks of length 12.7 cm (shown in Fig. 2) when one is hit "on-center" and the other is hit 3.4 cm to the right of center. The equation needed to calculate the rotational kinetic energy of rectangular block spinning abou its center of mass is E-410? [Eq. 1] 1 is the block's rotational inertia and w denotes the magnitude of its angular velocity. By using data on a typical block's mass and dimensions, you can easily calculate the values of I and W needed to determine it's rotational energy and to determine whether this energy is significant compared to the energy lost by the bullet when it's embedded in a block. The equation for rotational inertia, I derived in most introductory textbooks for a flat block of length L and width Wspinning around its center of mass is given by 1 = M(L + W2). [E9.23 The blocks used in this bullet-block experiment each had a mass, M, of about 0.222 kg and a fa length and width given by L = 0.127 m and W=0.102 m.
a) Rotational Inertia of the Spinning Block: Using data just presented for M, L and W, show that the block that was hit off-center had a rotational inertia of about 1=5.0 x10-kg-m. b) Rotational Speed of a Typical Spinning Block: To determine rotational speed, w in radians per second of a typical rotating block, we acquired typical spin rate data by viewing the video entitled <BB_Beta 1_Demo.mov>. One segment recorded at 2 min 52 sec shows a 500 fps B&W image of the spinning block that took 43 frames to complete a full turn. So the time for a single turn through an angle of 2 radians was given by T = (43/500)s = 0.086s per turn. Use the definition of w to show that the block's rotational speed in radians per second is 73 rad/s. W= c) Rotational Energy: Using Eq. 1 it can be shown that the block's rotational energy is only about 1.3 Joules. This is much smaller than the original kinetic energy of the bullet of about 170 Joules (as you calculated in Section III. (b)). d) Fraction of Bullet's Energy Transformed to the Kinetic Energy of the Block: The block being shot off-center appears to be spinning very rapidly. Even so, when a block hit off-center starts to rise, it only has about 1.3 Joules of rotational kinetic energy and 1.9 Joules of linear kinetic energy that you calculated in section III. d. This is a very small amount of energy compared to the 170 Joules of kinetic energy that the bullet had just before it entered the block! What percent of the bullet's initial KE contributed to the KE of the bullet block? V. Reflections on Your Findings Summarize what you learned about the nature of energy and momentum in non-elastic materials from viewing the bullet/block video for on-center and off-center bullets and completing this assignment. Use your own words to explain why blocks hit on-center and off-center seem to rise to the same height even though the block hit off center has additional rotational energy.