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ID: 1290050 • Letter: C
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could anyone help me how to solve this?
Edit View History Bookmarks Window Help Watch It Practice Another Pre Lecture10A-Chapter10-PHY2400 www.webassign.net/web/Student/Assignment-Responses/lastdep 9827081 Apple icloud Facebook Wikipedia Yahoo! News Popular Pre Lecture 10A-Chapter 10-PHY 2400 Chandra is correct. Note that because LCH for any of the objects in the simulation is some multiplicative factor of MR2, it is possible to calculate vcM knowing only this multiplicative factor and h Part 11 of 11 Analyze Now Chandra and Darcel decide to try a problem. suppose that the height of the incline is h 16.0 m. Find the speed at the bottom for each of the following objects solid sphere spherical shell cylinder In a race, which object would win? solid sphere spherical shell tieExplanation / Answer
at top PE = M*g*h
solid sphere
I = (2/5)*M*R^2
KEr = 0.5*I*W^2 + 0.5*M*V^2
KEr = 0.5*(2/5)*M*R^2*V^2/R^2 + 0.5*M*V^2
KEr = 0.5*(2/5)*M*V^2 + 0.5*M*V^2
KEr = 0.5*(7/5)*M*V^2
KEr = PE
0.5*(7/5)*M*V^2 = M*g*h
V = sqrt((10/7)*g*h) = 14.9 = 15 m/s
=----------------------
for spherical shell
I = (2/3)*M*R^2
KEr = 0.5*I*W^2 + 0.5*M*V^2
KEr = 0.5*(2/3)*M*R^2*V^2/R^2 + 0.5*M*V^2
KEr = 0.5*(2/3)*M*V^2 + 0.5*M*V^2
KEr = 0.5*(5/3)*M*V^2
KEr = PE
0.5*(5/3)*M*V^2 = M*g*h
V = sqrt((6/5)*g*h) = 13.71 m/s
====================
for hoop
I = *M*R^2
KEr = 0.5*I*W^2 + 0.5*M*V^2
KEr = 0.5*M*R^2*V^2/R^2 + 0.5*M*V^2
KEr = 0.5*M*V^2 + 0.5*M*V^2
KEr = M*V^2
KEr = PE
M*V^2 = M*g*h
V = sqrt(g*h) = 12.52 m/s
=======================
for cylinder
I = (1/2)*M*R^2
KEr = 0.5*I*W^2 + 0.5*M*V^2
KEr = 0.5*(1/2)*M*R^2*V^2/R^2 + 0.5*M*V^2
KEr = 0.5*(1/2)*M*V^2 + 0.5*M*V^2
KEr = 0.5*(3/2)*M*V^2
KEr = PE
0.5*(3/2)*M*V^2 = M*g*h
V = sqrt((4/3)*g*h) = 14.45 m/s
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