A small, solid sphere of mass 0.7 kg and radius 25 cm rolls without slipping alo

Question

A small, solid sphere of mass 0.7 kg and radius 25 cm rolls without slipping along the track consisting of slope and loop-the-loop with radius 3.3 m at the end of the slope. It starts from rest near the top of the track at a height h, where h is large compared to 25 cm. What is the minimum value of h (in terms of the radius of the loop R) such that the sphere completes the loop? The acceleration due to gravity is 9.8 m/s^2. The moment of inertia for a solid sphere is 2/5 m r^2. Answer in units of m. What are the force component in the horizontal direction on the sphere at the point p, which has coordinates (-R,0) if we take the center o the loop as origin, and if h = 3 R? Answer in units of N. The position vector of a particle of mass 2 kg is given as a function of time by = (7 m) + (5 m/s) t . Determine the magnitude of the angular momentum of the particle with respect to the origin at time 5 s. Answer in units of kg m^2/s.

Explanation / Answer

the minimum speed to complete the circle = v = sqrt(5*g*R)

PE at height h = M*g*h

KE at the bottom = 0.5*I*w^2 + 0.5*M*v^2

I = moment of inertia = (2/5)*M*r^2

w = v/r

KE = 0.5*(2/5)*M*v^2 + 0.5*M*V^2 = (10/7)*M*v^2 = (10/7)*M*5*g*R = (50/7)*g*R

from energy conservation

PE = KE

M*g*h = (50/7)*g*R

0.7*9.8*h = (50/7)*9.8*3.3

h = 33.67 m

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

KE at the bottom = (50/7)*g*R

at (-R,0)

KE = (50/7)*M*V1^2

PE = M*g*R

from energy conservation

M*g*h = M*g*R + (50/7)*M*V1^2

0.7*9.8*3*3.3 = (0.7*9.8*3.3)+((50/7)*0.7*v1^2)

v1 = 3 m/s

F = m*v1^2/R = 0.7*3^2/3.3 = 1.9 N

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

r = 7i+ 5t j

v = dr/dt = 5j m/s

at t = 5

r = 7i + 25 j

L = m*( r x v)

L = 2*( (7i + 25 j) x 5j )

L = 2*7*25 k

L = 350 kg m^2/s k

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