SOlutions to Aditional Problems

SOlutions to Aditional Problems - 236 Chapter 10*P10.54(a...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
236 Chapter 10 *P10.54 (a)± ±We±con±sid±er±the±el±e±va±tor-sheave-coun±ter±weight-Earth±sys±tem,±including± n passengers, as an iso±lat±ed±sys±tem±and±ap±ply±the±con±ser±va±tion±of±me±chan±i±cal±en±er±gy.±We±take±the±in±i±tial±con±F g- u±ra±tion,±at±the±mo±ment±the±drive±mech±a±nism±switches±off,±as±rep±re±sent±ing±ze±ro±grav±i±ta±tion±al± po±ten±tial±en±er±gy±of±the±sys±tem.± Therefore,±the±in±i±tial±me±chan±i±cal±en±er±gy±of±the±sys±tem±is E i = K i + U i = (1 / 2) m e v 2 + (1 / 2) m c v 2 + (1 / 2) I s ω 2 = (1 / 2) m e v 2 + (1 / 2) m c v 2 + (1 / 2)[(1 / 2) m s r 2 ]( v / r ) 2 = (1 / 2) [ m e + m c v 2 + (1 / 2) m s ] v 2 The F nal me chan i cal en er gy of the sys tem is en tire ly gravitational be cause the sys tem is mo±men±tar±i±ly±at±rest: E f = K f + U f = 0 + m e gd m c gd where we have rec og nized that the el e va tor car goes up by the same dis tance d that the coun ter weight goes down. Set ting the in i tial and F nal en er gies of the sys tem equal to each oth er, we have (1 / 2) [ m e + m c + (1 / 2) m s ] v 2 = ( m e m c ) gd (1 / 2) [800 kg + n 80 kg + 950 kg + 140 kg](3 m / s) 2 = (800 kg + n 80 kg 950 kg)(9.8 m / s 2 ) d d = [1890 + 80 n ](0.459 m) / (80 n – 150) (b) d = [1890 + 80 × 2](0.459 m) / (80 × 2 150) = 94.1 m (c) d = [1890 + 80 × 12](0.459 m) / (80 × 12 150) = 1.62 m (d) d = [1890 + 80 × 0](0.459 m) / (80 × 0 150) = 5.79 m (e) The rising car will coast to a stop only for n 2. ²or n = 0 or n = 1, the car would acceler- ate upward if released. (f ) The graph looks roughly like one branch of a hyperbola. It comes down steeply from 94.1 m for n = 2, fl attens out, and very slowly approaches 0.459 m as n becomes large. (g) The radius of the sheave is not necessary. It divides out in the expression (1/2) I ω 2 = (1/4) m sheave v 2 . (h) In this problem, as often in everyday life, energy conservation refers to minimizing use of electric energy or fuel. In physical theory, energy conservation refers to the constancy of the total energy of an isolated system, without regard to the different prices of energy in different forms. (i) The result of applying Σ F = ma and Σ τ = I α to elevator car, counterweight, and sheave, and adding up the resulting equations is (800 kg + n 80 kg – 950 kg)(9.8 m / s 2 ) = [800 kg + n 80 kg + 950 kg + 140 kg] a a = (9.80 m / s 2 )(80 n – 150) / (1 890 + 80 n ) downward
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Rotation of a Rigid Object About a Fixed Axis 237 *P10.55 (a) We mod el the as sem bly as a rig id body in equi lib ri um. Two torques acting on it are the fric tion al torque and the driv ing torque due to the emit ted wa ter: Σ τ = τ thrust friction = 0 3 F , b ω = 0 = 3 F , / b No tice that we have in clud ed a driv ing torque on±ly from the sin gle holes at dis tance , . Be cause of the third as sump tion, the ra di±al±ly-di±rect±ed±wa±ter±from±the±ends±ex±erts±no±torque± on the as sem bly—its thrust force is along the ra di al di rec tion. (b) We mod el the as sem bly as a rig id body un der a net torque. Be cause the as sem bly be gins from rest, there is no fric tion al torque at the be gin ning. Therefore, Σ = thrust = I α 3 F , = 3[ mL 2 / 3] = 3 F , / mL 2 (c) The con stant an gu lar speed with which the as sem bly ro tates will be larg er. The arms are
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 74

SOlutions to Aditional Problems - 236 Chapter 10*P10.54(a...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online