A hollow spherical shell with mass 1.65 kg rolls without slipping down a slope that makes an angle of 38.0 ∘ with the horizontal. Part A Find the magnitude of the acceleration acm of the center of mass of the spherical shell. Take the free-fall acceleration to be g = 9.80 m/s2 .

The spherical shell with mass m has moment of inertia I=2/3*m*R² Furthermore a pure rolling relates dω/dt and a through a = R dω/dt. So the two equations become

m*a = m*g sin(α) - F

2/3*m*a = F

IF we combine both:

m*a = m*g*sin(α) - 2/3*m*a

1.65a = 1.65*9.81 * sin(38.0) - 2/3 *1.65a

1.65a + 1.1a = 9.9654

2.75a = 9.9654

a = 3.62 m/s²

The acceleration is 3.62 m/s²

bhoopendrasisodiya34 bhoopendrasisodiya34 · Answer 2 · 2022-03-19T06:38:36+01:00

The magnitude of the acceleration of the center of mass of the spherical shell is 3.62 m/s².

What is Newton’s second law of motion?

Newton’s second law of motion shows the relation between the force mass and acceleration of a body. It says, that the net force applied on the body is equal to the product of mass of the body and the acceleration of it.

It can be given as,

[tex]\sum F=ma[/tex]

Here, (m) is the mass of the body and (a) is the acceleration.

A hollow spherical shell with mass 1.65 kg rolls without slipping down a slope that makes an angle of 38.0 ∘ with the horizontal. The free-fall acceleration g is 9.80 m/s2 .

The total friction force is equal to the force of gravity acting downward of the slope.

[tex]\sum F=mg\sin(\theta)-F\\[/tex]

For the force acting on the rotating spherical shell is,

[tex]F=\dfrac{2}{3}ma[/tex]

Put this value in above equation,

[tex]\sum F=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\ma=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\(1.65)a=(1.65)(9.80)\sin(38)-\dfrac{2}{3}(1.65)a\\a=(9.80)\sin(38)-\dfrac{2}{3}a\\a=3.62\rm \; m/s^2[/tex]

Thus, the magnitude of the acceleration of the center of mass of the spherical shell is 3.62 m/s².

Learn more about the Newton’s second law of motion here;

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