Answer:
Explanation:
Acceleration of a body rolling down an inclined plane is given by the following expression
a = g sinθ / 1 + k² / r²
θ is angle of inclination , k is radius of gyration and r is radius of the body rolling down .
for solid sphere
k² / r² = 2 / 5
= .4
a = g sinθ / 1 .4
= 7 sinθ
for spherical shell
k² / r² = 2 / 3
= .67
a = g sinθ / 1 .67
= 5.87 sinθ
for solid cylinder
k² / r² = 1/2
= .5
a = g sinθ / 1 .5
= 6.53 sinθ
for cylindrical shell
k² / r² = 1
a = g sinθ / 2
= 4.9 sinθ
Acceleration of solid cylinder is maximum ( 7 sinθ )
so it will reach bottom first.
Answer:
Solid Sphere
Explanation:
The object with the smallest moment of inertia will reach the bottom first.
Moment of inertia of the following:
Solid Sphere = 2/5 mr^2
Spherical Shell = 2/3 mr^2
Solid Cylinder = 1/2 mr^2
Cylindrical shell = mr^2
Since we know all the masses and radii are the same we can replace them with a placeholder value of mr^2 = 1 kgm^2 as an example.
Moment of inertia of the following (our example system):
Solid Sphere = 2/5 kgm^2
Spherical Shell = 2/3 kgm^2
Solid Cylinder = 1/2 kgm^2
Cylindrical shell = 1 kgm^2
We can see that the solid sphere will have the smallest moment of inertia therefore it will reach the bottom first.
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