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A uniform solid sphere rolls down an incline. (a) What must be the incline angle (deg) if the linear acceleration of the center of the sphere is to have a magnitude of 0.12g? (b) If a frictionless block were to slide down the incline at that angle, would its acceleration magnitude be more than, less than, or equal to 0.12g?

Respuesta :

Answer:

Explanation:

required acceleration = .12 g

acceleration of a rolling body on an inclined surface is given by the expression

= gsinθ / 1 + k² / R²

θ is the slope, k is radius of gyration of rolling body , R is radius of the rolling body .

Here rolling body is sphere

k² = 2/5 R²

k² / R² = 2/5

acceleration = g sinθ / 1 + k² / R²

.12 g = g sinθ / 1 + k² / R²

.12 g =g sinθ / 1 + 2/5

.12 = sinθ / 1.4

sinθ = .168

θ = 9.6 degree .

b ) If a frictional block falls down the incline , its acceleration will be

g sinθ = g sin 9.6

= 0.167 g . which is more than 0.12 g .

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