A hollow sphere of radius 0.23 m, with rotational inertia I = 0.034 kg m^2 about

Question

A hollow sphere of radius 0.23 m, with rotational inertia I = 0.034 kg m^2 about a line through its center of mass, rolls without slipping up a surface inclined at 29degree to the horizontal. At a certain initial position, the sphere's total kinetic energy is 21 J. How much of this initial kinetic energy is rotational? What is the speed of the center of mass of the sphere at the initial position? Now, the sphere moves 1.0 m up the incline from its initial position. What is its total kinetic energy now? What is the speed of its center of mass now? A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector = (2.00 m) - (3.00 m) + (2.00 m), the force is = F_x + (7.00 N) - (7.20 N) and the corresponding torque about the origin is = (7.60 N m) + (6.20 N m) + (1.70 N m). Determine F_x.

Explanation / Answer

total kinetic energy K = KEt + kEr

KEt = 0.5*m*v^2

KEr = 0.5*I*w^2

w = v/r

I = (2/3)*m*r^2

KEr = 0.5*(2/3)*m*v^2 = (1/3)*m*v^2

KE = KEt + KEr

KE = (5/6)*m*v^2

KEr/KE = (1/3)/(5/6) = 2/5

(b)

I = (2/3)*m*r^2

0.034 = (2/3)*m*0.23^2

m = 0.96 kg

KE = (5/6)*m*v^2

21 = (5/6)*0.96*v^2

v = 5.12 m/s

+++++++++++

(c)

KEi = KEf + m*g*L*sintheta

21 = KEf + (0.96*9.8*1*sin29)

KEf = 16.4 J

(d)

KEf = (5/6)*m*v^2

16.4 = (5/6)*0.96*v^2

v = 4.53 m/s

++++++++++++++++++++

torque = rxF

                             i      j      k

7.6i + 6.2 j + 1.7 k   =     2     -3      2

                             Fx    7     -7.2

7.6i + 6.2 j + 1.7 k =     7.6i + (14.4 + 2Fx)j + (14+3Fx) k

Fx = -4.1 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.