1. Bolo ea 0.2-kg, e hoketsoeng qetellong ea thapo e rapameng, e potoloha ka selikalikoe sa radius ea mithara e le 'ngoe 'me lebelo le phahameng ka ho fetisisa la bolo ke 10 rpm. Boholo ba ho potlaka ha centripetal le boholo ba matla a khatello ke bofe?
Tse tsejoang:
Boima (m) = 0.2 kg
Radius (r) = 1 m
Lebelo la Angular (ω) = 10 rev/min = 10 rev/60 s = 0.17 rev/s = (0.17)(6.28 rad)/s = 1 rad/s
Lebelo (v) = r ω = (1 m)(1 rad/s) = 1 m/s
Ho batloa: a s le Σ F
Tharollo:
(a) Boholo ba ho potlaka ha centripetal
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(b) Boholo ba matla a khatello
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. Bolo ea 1-kg qetellong ea khoele e potoloha ka ho lekana selikalikoeng se bataletseng sa radius ea 1 m. Thapo e tla robeha ha khatello e ho eona e feta 100 N. Lebelo le phahameng ka ho fetisisa leo bolo e ka bang le lona ke lefe?
Tse tsejoang:
Boima (m) = 1 kg
Radius (r) = 1 mithara
Matla a tsitsipano (T) = matla a bohareng ( Σ F) = 100 N
Ho batloa: v maximum
Tharollo:

[ID ea sephutheloana sa wpdm='499′]
- Boima le boima
- Matla a tloaelehileng
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- Matla a khohlano
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- Ho potoloha mothapo o kobehileng - matla a motsamao o chitja
- Motsamao o tšoanang ka selikalikoe se otlolohileng
- Matla a bohareng a motsamao o chitja o tšoanang