1. Bhora re 0.2-kg, rakasungirirwa kumagumo kwetambo yakatambanudzwa, rinotenderedzwa mudenderedzwa re radius yemita imwe uye kumhanya kukuru kwebhora i10 rpm. Chii chinonzi kukura kwekumhanya kwepakati uye kukura kwesimba rekumanikidza?
Zvinozivikanwa:
Huremu (m) = 0.2 kg
Nharaunda (r) = 1 m
Kumhanya kweAngular (ω) = 10 rev/min = 10 rev/60 s = 0.17 rev/s = (0.17)(6.28 rad)/s = 1 rad/s
Kumhanya (v) = r ω = (1 m)(1 rad/s) = 1 m/s
Zvaidiwa: a s uye Σ F
Solution:
(a) Hukuru hwekumhanya kwepakati pesimba
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(b) Hukuru hwesimba rekumanikidzwa
Σ F = ma
T = ma s
T = (0.2 kg)(1 m/s 2 )
T = 0.2 kg m/s 2
T = 0.2 N
2. Bhora re 1-kg riri kumagumo kwetambo riri kutenderera zvakaenzana mudenderedzwa rakati sandara re 1 m. Tambo yacho ichatyoka kana kumanikidzwa kuri mairi kwapfuura 100 N. Ndeipi kumhanya kwakanyanya kwaunogona kuita bhora?
Zvinozivikanwa:
Huremu (m) = 1 kg
Nharaunda (r) = mita imwe
Simba resimba (T) = simba repakati ( Σ F) = 100 N
Zvinodiwa: v maximum
Solution:

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