A uniform solid sphere of radius r=0.500 m and mass m=15.0 kg turns counterclockwise about a vertical axis through its centre. Find its vector angular momentum about this axis when its angular speed is 3.00 rad / s.

Question

A uniform solid sphere of radius r=0.500 m and mass m=15.0 kg turns counterclockwise about a vertical axis through its centre. Find its vector angular momentum about this axis when its angular speed is 3.00 rad / s. (A) (2.50 kg ⋅ m 2 / s ) hatk (B) -(4.50 kg ⋅ m 2 / s ) hatk (C) -(2.50 kg ⋅ m 2 / s ) hatk (D) (4.50 kg ⋅ m 2 / s ) hatk

Tardigrade Admin · Accepted Answer

The moment of inertia of the sphere about an axis through its centre is

I = \frac{2}{5} M R^{2} = \frac{2}{5} (15.0 k g) (0.500 m)^{2} = 1.50 k g \cdot m^{2}

Therefore, the magnitude of the angular momentum is

L = I ω = (1.50 k g \cdot m^{2}) (3.00 r a d / s) = 4.50 k g \cdot m^{2} / s

Since the sphere rotates counterclockwise about the vertical axis, the angular momentum vector is directed upward in the

+ z

direction. Thus,

L = (4.50 k g \cdot m^{2} / s) \hat{k}

Q. A uniform solid sphere of radius $r = 0.500 m$ and mass $m = 15.0 k g$ turns counterclockwise about a vertical axis through its centre. Find its vector angular momentum about this axis when its angular speed is $3.00 r a d / s$ .

$(2.50 k g \cdot m^{2} / s) \hat{k}$

$- (4.50 k g \cdot m^{2} / s) \hat{k}$

$- (2.50 k g \cdot m^{2} / s) \hat{k}$

$(4.50 k g \cdot m^{2} / s) \hat{k}$

Q. A uniform solid sphere of radius r=0.500m and mass m=15.0kg turns counterclockwise about a vertical axis through its centre. Find its vector angular momentum about this axis when its angular speed is 3.00rad/s.

(2.50kg⋅m2/s)k^

−(4.50kg⋅m2/s)k^

−(2.50kg⋅m2/s)k^

(4.50kg⋅m2/s)k^

Q. A uniform solid sphere of radius $r = 0.500 m$ and mass $m = 15.0 k g$ turns counterclockwise about a vertical axis through its centre. Find its vector angular momentum about this axis when its angular speed is $3.00 r a d / s$ .

$(2.50 k g \cdot m^{2} / s) \hat{k}$

$- (4.50 k g \cdot m^{2} / s) \hat{k}$

$- (2.50 k g \cdot m^{2} / s) \hat{k}$

$(4.50 k g \cdot m^{2} / s) \hat{k}$