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Answers to Chapter 7 problems

 

7.2.1   \(  \sqrt{2Rh-h^2}  \over R-h \)mg   

7.2.2   The container will slide before it tips over.   

7.2.3  (a)  7.0 m s-1    (b)  tan\( \theta \) = \(  vr  \over g^2 \)    (c)  45°     (d)   the pillion passenger will reduce the radius of curvature of the path and this will decrease vmin slightly.   

7.2.4  159 N;   137 N;   47 N

7.2.5  77°

7.6.1   \(  \Delta T  \over T \)  = \(  5h \rho  \over R_E \rho_E \)  

7.7.1  \(  3MR^2  \over 10\)

7.7.2  (a) 3.086 kg m2    (b) 6.38 kg m2     (i) 16%     (ii) 19%

7.7.3   1.46 x 10–46&²Ô²ú²õ±è;kg m2

7.7.4   2.82 x 10–3&²Ô²ú²õ±è;kg m2

7.7.5   0.73 kg m2

7.7.6  5.5 × 10 â€“34&²Ô²ú²õ±è;  J s

7.8.1  the length of the day increases by 0.81 s

7.8.2  \(  1\over 4 \)mg

7.8.3  (a) 3.14 rad s â€“1    (b)  4.23 rad s–1     (c)  1515 J;  1459 J 

7.8.4   0.14 rad s–1

7.8.5  9.5 x 1041 J s;  0.4 ms

7.8.6  (a) 463 rad s–2     (b)  556 rad s–1

7.9.1  (a) 15.2 rad s–1    (b) 0.355 kg m2 s–1    (c) 13.1 rad s–2 

7.9.2  2.70 m;   â€“29.6 rad s–2&²Ô²ú²õ±è;&²Ô²ú²õ±è;

7.9.3  (a) 2.82 m s–1   (b) 17.6 rad s–1   (c)  3.17 kg m2 s–1

7.9.4  (a) 0.11 kg m2 s–1    (b) 1.73 J    (c) â€“12.1 rad s–2    (d)  0.042 N m  (e)  1.73 J

7.9.5   \(  a  \over {\sqrt{2}} \);    0.176 m

7.9.6   \(  2 \pi  \sqrt{16R^2 + 25l^2 \over 30lg} \);  R = 0.27 m

7.9.7   hoop:  5.8 m s–1;    cylinder:  6.7 m s–1;   sphere:  6.9 m s–1

7.9.8  1.7 s

7.10.1   5.8 × 10–10  m N rad–1 

7.12.1   \(  2 \pi  \sqrt{ \frac{3}{2} M \over k} \)

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