Answer:
(a) 12.25 Ā°
(b) The acceleration magnitude will be more than 0.063Ā·g or 0.212Ā·g
Explanation:
(a) Here we have
V = u + at
And KE = KE[tex]_R[/tex] + KE[tex]_T[/tex] = 0.5 mĀ·vĀ² + 0.5 IĻĀ²
I = 2/3mĀ·rĀ², Ļ = v/r
KE = 0.5 mĀ·vĀ² + 2/3 mĀ·rĀ²Ā·v/r
Therefore = 0.5 mĀ·vĀ² + 2/3 mĀ·rĀ·v
Therefore, v is reduced by
v = ā(gĀ·hĀ·10/7) or vĀ² = gĀ·hĀ·10/7 compared to vĀ² = 2gh
If a = 0.063 g we have
vĀ² = 0.063Ā·gĀ·hĀ·10/7 = 0.09Ā·gĀ·h hence the height which is
v = 0.3ā(gĀ·h) and Ā v = ā(2Ā·gĀ·h)
Therefore, the angle is
sinā»Ā¹ (0.3/ā2) = 12.25 Ā°
v = u + at where
h = ut + 0.5atĀ²
(b) At that angle, we have
hā = 0.5gtĀ² and
hā = Ā 0.212 Ć hā
Therefore
a Ā = 0.212 g, the acceleration will be more than 0.063Ā·g.
Answer:
the inclination of the incline angle Īø = 5.06Ā°
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]
ā“
[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 Īø
sin Īø = [tex]\frac{0.063}{0.7143}[/tex]
sin Īø = 0.0882
Īø = sinā»Ā¹ (0.0882)
Īø = 5.06Ā°
Thus; the inclination of the incline angle Īø = 5.06Ā°
b) From the question ; the block is said to be friction-less , therefore with that, the acceleration magnitude Ā remains the same since it contains no friction related terms.
