Answer:
(a) 12.25 °
(b) The acceleration magnitude will be more than 0.063·g or 0.212·g
Explanation:
(a) Here we have
V = u + at
And KE = KE[tex]_R[/tex] + KE[tex]_T[/tex] = 0.5 m·v² + 0.5 Iω²
I = 2/3m·r², ω = v/r
KE = 0.5 m·v² + 2/3 m·r²·v/r
Therefore = 0.5 m·v² + 2/3 m·r·v
Therefore, v is reduced by
v = √(g·h·10/7) or v² = g·h·10/7 compared to v² = 2gh
If a = 0.063 g we have
v² = 0.063·g·h·10/7 = 0.09·g·h hence the height which is
v = 0.3√(g·h) and v = √(2·g·h)
Therefore, the angle is
sin⁻¹ (0.3/√2) = 12.25 °
v = u + at where
h = ut + 0.5at²
(b) At that angle, we have
h₁ = 0.5gt² and
h₂ = 0.212 × h₁
Therefore
a = 0.212 g, the acceleration will be more than 0.063·g.
Answer:
the inclination of the incline angle θ = 5.06°
the acceleration magnitude is said to remains the same
Explanation:
The acceleration of the uniform solid sphere down the incline can be illustrated as:
[tex]a = \frac{gsin \theta }{1 + \frac{k^2}{r^2}}[/tex]
where k = radius of gyration ; and for solid sphere ; k = [tex](\frac{2}{5})^2[/tex]
∴
[tex]a = \frac{gsin \theta }{1 + \frac{\frac{2}{5}r^2 }{r} }[/tex]
[tex]a = (\frac{5}{7})g \ sin \theta[/tex]
If the linear acceleration of the center of the sphere is to have a magnitude of 0.063 g; Then; we have:
0.063 g = [tex](\frac{5}{7})g \ sin \theta[/tex]
0.063 = 0.7143 sin θ
sin θ = [tex]\frac{0.063}{0.7143}[/tex]
sin θ = 0.0882
θ = sin⁻¹ (0.0882)
θ = 5.06°
Thus; the inclination of the incline angle θ = 5.06°
b) From the question ; the block is said to be friction-less , therefore with that, the acceleration magnitude remains the same since it contains no friction related terms.
