Đáp án:
Giải:
Ta có:
`mgl(1-cos\alpha_0)=\frac{1}{2}mv^2+mgl(1-cos\alpha)`
`2mgl(1-cos\alpha_0)=mv^2+2mgl(1-cos\alpha)`
`2gl(1-cos\alpha_0)=v^2+2gl(1-cos\alpha)`
`v^2=2gl(1-cos\alpha_0)-2gl(1-cos\alpha)`
`v^2=2gl(1-cos\alpha_0-1+cos\alpha)`
`v^2=2gl(cos\alpha-cos\alpha_0)`
Do đó:
`T=ma_{ht}+P_1=m\frac{v^2}{r}+P_1=m\frac{v^2}{l}+P_1`
`T=m\frac{2gl(cos\alpha-cos\alpha_0)}{l}+mgcos\alpha`
`T=2mg(cos\alpha-cos\alpha_0)+mgcos\alpha`
`T=mg(2cos\alpha-2cos\alpha_0+cos\alpha)`
`T=mg(3cos\alpha-2cos\alpha_0) \ (N)`
