How Do You Calculate Acceleration and Friction for a Rolling Spherical Shell?

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Homework Help Overview

The discussion revolves around calculating the acceleration, friction force, and minimum coefficient of friction for a hollow spherical shell rolling down an inclined plane at a specific angle. The problem involves concepts from dynamics and energy conservation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between potential and kinetic energy for the rolling shell, with attempts to derive acceleration and friction force using energy conservation principles and Newton's second law. There are questions about specific calculations and assumptions regarding the angle used in trigonometric functions.

Discussion Status

Some participants are verifying calculations and questioning the correctness of trigonometric evaluations. There is an ongoing exploration of the implications of using different units for angle measurement, particularly in relation to the sine function.

Contextual Notes

Participants note discrepancies in calculated values and the importance of ensuring the correct unit system is used for angle measurements. There is an emphasis on checking assumptions related to the rolling condition and the energy balance.

Black_Hole???
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Q1.) A hollow, spherical shell with mass 1.70 rolls without slipping down a slope angled at 38.0.

-Find the acceleration.
-Find the friction force.
-Find the minimum coefficient of friction needed to prevent slipping.

If it rolls wthout slippng, the increase in Kinetic Energy equals the decrease in Potential Energy. The kinetic energy is in two parts, rotation and translation.
M g H sin 38 = (1/2) M V^2 + (1/2) I (V/R)^2
= [(1/2) + (1/3)] M V^2
The M's and R's cancel and
gH sin 38 = (5/6) V^2
V = sqrt (1.2 g H sin 38) = sqrt (2 a H)
where a is the acceleration
a = 0.6 g sin 38

= 1.74 which is wrong



Use that acceleration and Newton's second law t compute the actual friction force, F.
M g sin 38 - F = M a
Mg sin 38 - M*0.6 g sin 38 = F
F = 0.4 M g sin 38

=1.975 which is wrong

To provide this amount of friction, the static coefficient of friction mu,s must equal or exceed a value given by
M g cos 38 * mu,s = 0.4 M g sin 38
mu,s = 0.4 tan 38

=.124 which is also wrong.

Can someone please check over what i did and see if i miscalculated or something.
 
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Black_Hole? said:
Q1.) A hollow, spherical shell with mass 1.70 rolls without slipping down a slope angled at 38.0.

-Find the acceleration.
-Find the friction force.
-Find the minimum coefficient of friction needed to prevent slipping.

If it rolls wthout slippng, the increase in Kinetic Energy equals the decrease in Potential Energy. The kinetic energy is in two parts, rotation and translation.
M g H sin 38 = (1/2) M V^2 + (1/2) I (V/R)^2
= [(1/2) + (1/3)] M V^2
The M's and R's cancel and
gH sin 38 = (5/6) V^2
V = sqrt (1.2 g H sin 38) = sqrt (2 a H)
where a is the acceleration
a = 0.6 g sin 38

= 1.74 which is wrong

6/5*g*x*sin38 = 2*a*x

a = .6*g*sin38 = .6*9.8*.616 = 3.63 m/s2
 
ok how does the sin(38) = .616?

i keep getting .296369

It is the right answer though
 
You're evaluating the sine in units of radians. Switch to degrees (or multiply 38 with pi/180 before evaluating the sine).
 

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