Question

1) A torque of 1.20 N m is applied to a thin rod of mass 2.50 kg and length 50.0 cm pivoted about its center and at rest. How fast is the rod spinning after 4.25 s?

a. 32.6 rad/s

b. 8.16 rad/s

c. 97.9 rad/s

d. 24.5 rad/s

2) A torque of 1.20 N m is applied to a thin rod of mass 2.50 kg and length 50.0 cm pivoted about one end and at rest. How fast is the rod spinning after 4.25 s?

a.		8.16 rad/s
b.		24.5 rad/s
c.		97.9 rad/s
d.		32.6 rad/s

3) A figure skater jumps into the air with their arms outstretch and begins rotating at 1.6 rev/s. They pull in their arms reducing their moment of inertia to 2/7 its original value. What is their new rate of rotation?

a.		0.86 rev/s
b.		3.0 rev/s
c.		5.6 rev/s
d.		0.46 rev/s

4) A person of mass m = 85.0 kg stands at the edge of a large disk of mass M = 125 kg and radius R = 15.0 m rotating with an angular speed of 0.320 rad/s about an axis through its center. The person begins walking straight toward the center of the disk. How fast is the disk spinning when the person is 3.50 m from the disk's center?

a		0.518 rad/s
b		0.486 rad/s
c		0.320 rad/s
d		0.740 rad/s

5) A projectile of mass m = 15.0 g traveling horizontally at 175 m/s strikes and sticks to (embeds in) a vertical rod of mass 125 g and length L = 1.40 m pivoted about the center. Determine the angular speed of the rod after the collision if the projectile sticks a distance d = 50.0 cm from the center of the rod.

	a	64.3 rad/s
	b	108 rad/s
	c	54.3 rad/s
	d	138 rad/s

6) A 5.0 kg particle located at the position (2.0 m, -1.0 m, 3.0 m) has a velocity given by v = (3.0 m/s)i + (2.0 m/s)j - (1.0 m/s)k. What is its angular momentum about the origin?

a		(25 kg m2/s)i - (55 kg m2/s)j - (35 kg m2/s)k
b		-(25 kg m2/s)i + (55 kg m2/s)j + (35 kg m2/s)k
c		(30 kg m2/s)i - (10 kg m2/s)j - (15 kg m2/s)k
d		(15 kg m2/s)i + (10 kg m2/s)j - (5 kg m2/s)k

7) A 5.0 kg particle located at the position (2.0 m, -1.0 m, 3.0 m) has a velocity given by v = (3.0 m/s)i + (2.0 m/s)j - (1.0 m/s)k. What is the magnitude of its angular momentum about the origin?

a		5.0 kg m2/s
b		18.7 kg m2/s
c		35 kg m2/s
d		70 kg m2/s

Answer 1

(1) Moment of inertia of the rod about its center = I = ML² / 12,

where, M = 2.5 kg is mass of the rod, L = 50 cm = 0.5 m is length of the rod. hence,

Hence, torque applied = I = 1.2, in N . m,

where, is the angular acceleration of the rod.

Hence, = 1.2 / I = ( 1.2 x 12 ) / ML² = ( 1.2 x 12 ) / ( 2.5 x 0.5² )

or, = 23.04 rad / s².

Initial angular velocity = _o = 0 rad / s.

Hence, angular velocity after time t = 4.25 s is : = _o + t = 0 + 23.04 x 4.25 rad / s

or, = 97.9 rad / s.

Hence, correct option is : (c) 97.9 rad / s.